\section{Monoidal structures and other operations} \label{sec:monoidal}

In this section, we study operations on constructible directed complexes and their extension to constructible polygraphs. Some of them are the same that Steiner considered in \cite[Theorem 2.19]{steiner1993algebra} for directed complexes.

\begin{cons}
Let $P, Q$ be oriented graded posets. The disjoint union $P + Q$ of $P$ and $Q$ as posets is graded, and inherits an orientation from its components. Clearly, $P + Q$ is an oriented thin poset or a constructible directed complex whenever $P$ and $Q$ are.
\end{cons}

\begin{cons} \label{cons:suspension}
Let $P$ be an oriented graded poset. The \emph{suspension} $\Sigma P$ of $P$ is the oriented graded poset obtained from $P$ by adjoining two minimal elements $\bot^-$ and $\bot^+$, such that $\bot^\alpha < x$ for all $x \in P$, and, for all 0-dimensional $y \in P$, $o(c_{y,\bot^-}) := -$ and $o(c_{y,\bot^+}) := +$. If $\dmn{x} = n$ in $P$, then $\dmn{x} = n+1$ in $\Sigma P$.

This construction extends to an endofunctor $\Sigma$ on $\ogpos$ in the obvious way.
\end{cons}

\begin{lem}
If $P$ is an oriented thin poset, then $\Sigma P$ is an oriented thin poset.
\end{lem}
\begin{proof}
Let $[x,y]$ be an interval of length 2 in $\Sigma P_\bot$.
\begin{itemize}
	\item If $\dmn{x} > 0$ in $\Sigma P$, then $[x,y]$ corresponds to an interval in $P$, which is of the required form (\ref{eq:signed}) by hypothesis.
	\item If $\dmn{x} = 0$ in $\Sigma P$, then $x = \bot^\alpha$, and by construction $[\bot^\alpha,y]$ is isomorphic to the interval $[\bot,y]$ in $P_\bot$ as a poset. By oriented thinness of $P$, it is of the form
\begin{equation*}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (0,2) {$y$};
	\node[scale=1.25] (1) at (-1,1) {$z_1$};
	\node[scale=1.25] (1b) at (1,1) {$z_2$};
	\node[scale=1.25] (2) at (0,0) {$\bot^\alpha$};
	\draw[1c] (0) to node[auto,swap] {$\beta$} (1);
	\draw[1c] (0) to node[auto] {$-\beta$} (1b);
	\draw[1c] (1) to node[auto,swap] {$\alpha$} (2);
	\draw[1c] (1b) to node[auto] {$\alpha$} (2);
	\node[scale=1.25] at (1.25,0) {$.$};
\end{tikzpicture}
\end{equation*}
	\item If $x = \bot$, then $y$ is a 0-dimensional element of $P$, and by construction $[\bot,y]$ is
	\begin{equation*}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (0,2) {$y$};
	\node[scale=1.25] (1) at (-1,1) {$\bot^-$};
	\node[scale=1.25] (1b) at (1,1) {$\bot^+$};
	\node[scale=1.25] (2) at (0,0) {$\bot$};
	\draw[1c] (0) to node[auto,swap] {$-$} (1);
	\draw[1c] (0) to node[auto] {$+$} (1b);
	\draw[1c] (1) to node[auto,swap] {$+$} (2);
	\draw[1c] (1b) to node[auto] {$+$} (2);
	\node[scale=1.25] at (1.25,0) {$.$};
\end{tikzpicture}
\end{equation*}
\end{itemize}
This proves that $\Sigma P$ is an oriented thin poset.
\end{proof}

\begin{prop} \label{prop:suspensionglob}
Let $P$ be an oriented thin poset and $U$ a closed subset. Then $U$ is a constructible $(n-1)$\nbd molecule in $P$ if and only if $\Sigma U$ is a constructible $n$\nbd molecule in $\Sigma P$. Consequently, if $P$ is a constructible directed complex, then $\Sigma P$ is a constructible directed complex.
\end{prop}
\begin{proof}
For all $0$\nbd dimensional $x \in P$, $\clos{\{x\}}$ is isomorphic to the 1-globe in $\Sigma P$. For all other 0-dimensional $y$ in $P$, it is always the case that $\bord{}{\alpha}x \cap \bord{}{-\alpha}y$ is empty in $\Sigma P$; it follows that there are no non-atomic 1\nbd molecules in $\Sigma P$. Hence, we can establish a bijection between $\glob{1}{(\Sigma P)}$ and $\glob{0}{P}$, and use it as the basis of an inductive proof that $\glob{n}{(\Sigma P)}$ is in bijection with  $\glob{n-1}{P}$, for all $n > 0$: if $U$ is a constructible $(n-1)$\nbd molecule in $P$, then $\Sigma U = U \cup \{\bot^+,\bot^-\}$ is a constructible $n$\nbd molecule in $\Sigma P$.

Because $\Sigma \clos{\{x\}}$ is just $\clos\{x\}$ in $\Sigma P$, it follows that $\Sigma P$ is a constructible directed complex whenever $P$ is.
\end{proof}

\begin{exm}
For all $n > 0$, $\Sigma (O^{n-1})$ is isomorphic to $O^n$.
\end{exm}

\begin{cons} \label{cons:laxgray}
Let $P, Q$ be oriented graded posets. The \emph{lax Gray product} $P \tensor Q$ of $P$ and $Q$ is the product poset $P \times Q$, which is graded, with the following orientation: write $x \tensor y$ for an element $(x,y)$ of $P \times Q$, seen as an element of $P \tensor Q$; then, for all $x'$ covered by $x$ in $P$, and all $y'$ covered by $y$ in $Q$, let
\begin{align*}
	o(c_{x \tensor y, x' \tensor y}) & := o_P(c_{x,x'}), \\
	o(c_{x \tensor y, x \tensor y'}) & := (-)^{\dmn{x}}o_Q(c_{y,y'}),
\end{align*}
where $o_P$ and $o_Q$ are the orientations of $P$ and $Q$, respectively.

The lax Gray product defines a monoidal structure on $\ogpos$, whose monoidal unit is $1$, the oriented graded poset with a single element.
\end{cons}

\begin{remark}
The choice of orientation comes from tensor products of chain complexes; we will see from Construction \ref{cons:adc} that it ensures that a certain functor from $\globpos$ to a category of chain complexes with additional structure is monoidal.
\end{remark}

\begin{lem} \label{lem:thintensor}
If $P, Q$ are oriented thin posets, then so is $P \tensor Q$.
\end{lem}
\begin{proof}
An interval of length 2 in $P \tensor Q$ has one of the following forms:
\begin{itemize}
	\item $[x', x] \tensor \{y\}$ for some interval $[x', x]$ of length 2 in $P$, and $y \in Q$;
	\item $\{x\} \tensor [y',y]$ for some interval $[y',y]$ of length 2 in $Q$, and $x \in P$;
	\item $[x'\tensor y', x \tensor y]$, where $x$ covers $x'$ in $P$ and $y$ covers $y'$ in $Q$.
\end{itemize}
In the first two cases, oriented thinness follows from the oriented thinness of $P$ and $Q$, respectively. In the last case, let $\alpha := o_P(c_{x,x'})$, $\beta := o_Q(c_{y,y'})$. Then the interval has the form
\begin{equation*}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}, scale=1.5]
	\node[scale=1.25] (0) at (0,2) {$x\tensor y$};
	\node[scale=1.25] (1) at (-1,1) {$x' \tensor y$};
	\node[scale=1.25] (1b) at (1,1) {$x \tensor y'$};
	\node[scale=1.25] (2) at (0,0) {$x' \tensor y'$};
	\draw[1c] (0) to node[auto,swap] {$\alpha$} (1);
	\draw[1c] (0) to node[auto] {$(-)^{\dmn{x}}\beta$} (1b);
	\draw[1c] (1) to node[auto,swap] {$(-)^{\dmn{x'}}\beta$} (2);
	\draw[1c] (1b) to node[auto] {$\alpha$} (2);
\end{tikzpicture}
\end{equation*}
in the labelled Hasse diagram of $P \tensor Q$, and $(-)^{\dmn{x'}} = (-)^{\dmn{x} - 1} = - (-)^{\dmn{x}}$. The case of intervals $[\bot, x \tensor y]$ in $(P \tensor Q)_\bot$ can easily be handled explicitly.
\end{proof}

In fact, the lax Gray product preserves the class of constructible molecules, and consequently of constructible directed complexes. We give a first sketch of the proof.
\begin{lem} \label{lem:globtensor}
Let $U$ be a constructible $n$\nbd molecule and $V$ a constructible $m$\nbd molecule. Then $U \tensor V$ is a constructible $(n+m)$\nbd . If $U' \sqsubseteq U$ and $V' \sqsubseteq V$, then $U' \tensor V' \sqsubseteq U \tensor V$.
\end{lem}
\begin{proof}[Sketch of the proof]
We proceed by double induction on the dimension and number of maximal elements of $U$ and $V$. If $U$ or $V$ is 0-dimensional, then $U \tensor V$ is isomorphic to $V$ or $U$, respectively, and there is nothing to prove. 

Suppose $n, m > 0$; then $U \tensor V$ is pure and $(n+m)$\nbd dimensional. We first need to show that $\bord{}{\alpha}(U \tensor V)$ is a constructible molecule. We have
\begin{equation*}
	\bord{}{\alpha}(U \tensor V) = (\bord{}{\alpha}U \tensor V) \cup (U \tensor \bord{}{(-)^n\alpha}V);
\end{equation*}
by the inductive hypothesis, both $(\bord{}{\alpha}U \tensor V)$ and $(U \tensor \bord{}{(-)^n\alpha}V)$ are constructible, and their intersection $\bord{}{\alpha}U \tensor \bord{}{(-)^n\alpha}V$ is constructible. We can show that this is, in fact, a decomposition into constructible submolecules.

If $U$ and $V$ are atomic, there is nothing else to prove. Suppose $U$ is non-atomic and splits as $U_1 \cup U_2$, so $U \tensor V = (U_1 \tensor V) \cup (U_2 \tensor V)$. By the inductive hypothesis, $(U_1 \tensor V)$ and $(U_2 \tensor V)$ are both constructible $(n+m)$\nbd molecules, and their intersection $(U_1 \tensor V) \cap (U_2 \tensor V) = (U_1 \cap U_2) \tensor V$ is a constructible $(n+m-1)$\nbd molecule. We can show that this is a decomposition into constructible submolecules.

In the same way, if $V$ splits as $V_1 \cup V_2$, then $U \tensor V$ splits as $(U \tensor V_1) \cup (U \tensor V_2)$.
\end{proof}

While the structure of the proof is straightforward, checking the various submolecule conditions on the boundaries, or that the $\bord{}{\beta}(\bord{}{\alpha}(U \tensor V))$ are themselves submolecules, requires a series of tedious inductive arguments. We give a more detailed proof in Appendix \ref{sec:appendix}.

\begin{prop} \label{thm:globpostensor}
If $P, Q$ are constructible directed complexes, then so is $P \tensor Q$.
\end{prop}
\begin{proof}
For all $x \in P$ and $y \in Q$, we have $\clos\{x \tensor y\} = \clos\{x\} \tensor \clos\{y\}$, which is a constructible molecule by Lemma \ref{lem:globtensor}.
\end{proof}

Thus, the monoidal structure $(\ogpos, \boxtimes, 1)$ restricts to $\globpos$.

\begin{exm}
The $n$\nbd fold lax Gray product $K^n := \underbrace{\vec{I}\tensor\ldots\tensor\vec{I}}_n$ is the $n$\nbd cube, with the conventional orientation of homological algebra \cite{al2002multiple}.
\end{exm}

\begin{cons} \label{cons:join}
Let $P, Q$ be oriented graded posets, and take the lax Gray product $P_\bot \tensor Q_\bot$. This has a unique 0-dimensional element $\bot \tensor \bot$, and all 1-dimensional elements $x \tensor \bot$ and $\bot \tensor y$ cover it with orientation $+$; thus, $P_\bot \tensor Q_\bot$ is isomorphic to $(P \join Q)_\bot$ for a unique oriented graded poset $P \join Q$, the \emph{join} of $P$ and $Q$. 

We introduce the following notation for elements of $P \join Q$:
\begin{itemize}
	\item for all $x \in P^{(n)}$, we write $x \in P \join Q$ for the $n$\nbd dimensional element corresponding to $x \tensor \bot$ in $P_\bot \tensor Q_\bot$;
	\item for all $y \in Q^{(m)}$, we write $y \in P \join Q$ for the $m$\nbd dimensional element corresponding to $\bot \tensor y$ in $P_\bot \tensor Q_\bot$;
	\item for all $x \in P^{(n)}$, $y \in Q^{(m)}$, we write $x \join y \in P \join Q$ for the $(n+m+1)$\nbd dimensional element corresponding to $x \tensor y$ in $P_\bot \tensor Q_\bot$.
\end{itemize}
The first two define inclusions $P \hookrightarrow P \join Q$ and $Q \hookrightarrow P \join Q$. 

The join is an associative operation because the lax Gray product is, and it extends to a monoidal structure on $\ogpos$, whose unit is the empty oriented graded poset $\emptyset$. This makes $(-)_\bot$ a monoidal functor from $(\ogpos, \join, \emptyset)$ to $(\ogpos, \tensor, 1)$.
\end{cons}

\begin{remark}
By monoidal functor, we always mean a \emph{strong} monoidal functor.
\end{remark}

\begin{lem}
If $P, Q$ are oriented thin posets, then so is $P \join Q$.
\end{lem}
\begin{proof}
An interval of length 2 in $(P \join Q)_\bot = P_\bot \tensor Q_\bot$ is either of the form $[x',x]\tensor\{y\}$, or of the form $\{x\}\tensor [y',y]$, or of the form $[x'\tensor y',x \tensor y]$ for some $x', x \in P_\bot$ and $y', y \in Q_\bot$. In the first two cases, we can use oriented thinness of $P$ and $Q$. In the third case, if $x', y' \neq \bot$ we can use oriented thinness of $P \tensor Q$. The remaining few cases are easily checked explicitly.
\end{proof}

\begin{prop} \label{lem:globjoin}
Let $U, V$ be constructible molecules, $\dmn{U} = n, \dmn{V} = m$. Then $U \join V$ is a constructible $(n+m+1)$\nbd molecule. If $U' \sqsubseteq U$ and $V' \sqsubseteq V$, then $U' \join V' \sqsubseteq U \join V$.
\end{prop}
\begin{proof}
Consider the following assignment $j$ of elements of $\Sigma U \tensor \Sigma V$ to elements of $U \join V$ (not an inclusion of oriented graded posets):
\begin{equation*}
	j(z) := \begin{cases} x \tensor \bot^+ & \text{if } z = x \in P, \\
		\bot^+ \tensor y & \text{if } z = y \in Q, \\
		x \tensor y & \text{if } z = x \join y, x \in P, y \in Q.
		\end{cases}
\end{equation*}
This is induced by $(U \join V)_\bot = (U_\bot \tensor V_\bot)$ via the functions $U_\bot \to \Sigma U$, $V_\bot \to \Sigma V$ that map $\bot$ to $\bot^+$. 

By Proposition \ref{prop:suspensionglob}, $U' \sqsubseteq U$ and $V' \sqsubseteq V$ imply $\Sigma U' \sqsubseteq \Sigma U$ and $\Sigma V' \sqsubseteq \Sigma V$. Furthermore, $\bord{}{\alpha}(\Sigma U) = \Sigma \bord{}{\alpha} U$ when $U$ is at least 1-dimensional, and $\bord{}{\alpha}(\Sigma U) = \{\bot^\alpha\}$ when it is 0-dimensional. 

Using these facts, it is straightforward to turn the inductive proof that $\Sigma U \tensor \Sigma V$ is a constructible $(n+m+2)$\nbd molecule, as in Lemma \ref{lem:globtensor}, and $\Sigma U' \tensor \Sigma V' \sqsubseteq \Sigma U \tensor \Sigma V$, into a proof that $U \join V = \invrs{j}(\Sigma U \tensor \Sigma V)$ is a constructible $(n+m+1)$\nbd molecule, and $U' \join V' = \invrs{j}(\Sigma U' \tensor \Sigma V') \sqsubseteq U \join V$. If $n = m = 0$, then we can directly check that $U \join V$ is isomorphic to $O^1$. If $n = 0, m > 0$, then
\begin{equation*}
	\bord{}{\alpha}(U \join V) = \invrs{j}\bord{}{\alpha}(\Sigma U \tensor \Sigma V) = \invrs{j}(\{\bot^\alpha\} \tensor \Sigma V \cup \Sigma U \tensor \Sigma \bord{}{-\alpha}V),
\end{equation*}
hence
\begin{equation*}
	\bord{}{+}(U \join V) = V \cup (U \join \bord{}{-}V), \quad \quad \bord{}{-}(U \join V) = U \join \bord{}{+}V,
\end{equation*}
where, $V, U \join \bord{}{-}V$ and $U \join \bord{}{+}V$ are all constructible $m$\nbd molecules, and the intersection $V \cap U \join \bord{}{-}V = \bord{}{-}V$ is a constructible $(m-1)$\nbd molecule; we can show that $V \cup (U \join \bord{}{-}V)$ is a decomposition into constructible submolecules. The case $n >0, m = 0$ is dual, and for $n > 0, m > 0$, or when $U$ or $V$ are not atoms, one proceeds exactly as in Lemma \ref{lem:globtensor}.
\end{proof}

\begin{prop}
If $P, Q$ are constructible directed complexes, then so is $P \join Q$.
\end{prop}
\begin{proof}
For all $x \in P$ and $y \in Q$, that $\clos\{x\}$ and $\clos\{y\}$ are constructible molecules in $P \join Q$ is immediate from the existence of inclusions $P, Q \hookrightarrow P \join Q$, and $\clos\{x \join y\} = \clos\{x\} \join \clos\{y\}$ is a constructible molecule by Proposition \ref{lem:globjoin}.
\end{proof}

It follows that the monoidal structure $(\ogpos, \join, \emptyset)$ also restricts to $\globpos$; notice however that $(-)_\bot$ is not an endofunctor on $\globpos$.

\begin{exm}
The $(n+1)$\nbd fold join $\Delta^n := \underbrace{1\join\ldots\join 1}_{n+1}$ is the oriented $n$\nbd simplex \cite{street1987algebra}.
\end{exm}

The lax Gray product and join of oriented graded posets are related to the tensor product of chain complexes, and Steiner's extension to augmented directed complexes, in the following way.

\begin{dfn}
Let $K$ and $L$ be two chain complexes. The \emph{tensor product} $K \otimes L$ of $K$ and $L$ is the chain complex defined by
\begin{equation*}
	(K \otimes L)_n := \bigoplus_{k=0}^n K_k \otimes L_{n-k}, 
\end{equation*}
\begin{equation*}
	d(x \otimes y) := dx \otimes y + (-1)^{\dmn{x}} x \otimes dy,
\end{equation*}
where $\dmn{x} = n$ if $x \in K_n$. If $K$ and $L$ are augmented, $K \otimes L$ is augmented by $e: x \otimes y \mapsto ex \, ey$.

If $K$ is an augmented chain complex, the \emph{suspension} of $K$ is the chain complex $\Sigma K$ defined by
\begin{equation*}
	(\Sigma K)_n := \begin{cases} \mathbb{Z}, & n = 0, \\ K_{n-1}, & n > 0, \end{cases} 
\end{equation*} 
\begin{equation*}
	(d: \Sigma K_{n+1} \to \Sigma K_n) := \begin{cases} e: K_0 \to \mathbb{Z}, & n = 0, \\ d: K_{n} \to K_{n-1}, & n > 0. \end{cases}
\end{equation*}
Because $(\Sigma K \otimes \Sigma L)_0 = \mathbb{Z} \otimes \mathbb{Z} \simeq \mathbb{Z}$, there is a unique (up to isomorphism) augmented chain complex $K \join L$ such that $\Sigma(K \join L) \simeq \Sigma K \otimes \Sigma L$. This augmented chain complex is the \emph{join} of $K$ and $L$.

Let $(K, K^*)$ and $(L, L^*)$ be augmented directed complexes. The \emph{tensor product} $(K, K^*) \otimes (L, L^*)$ is $K \otimes L$ together with the distinguished submonoids $(K \otimes L)_n^*$ generated by elements $x \otimes y$ of $(K \otimes L)_n$ where $x \in K_k^*$ and $y \in L_{n-k}^*$. 

The \emph{suspension} $\Sigma(K, K^*)$ is $\Sigma K$ with the distinguished submonoids $\mathbb{N}$ of $(\Sigma K)_0$ and $K^*_{n-1}$ of $(\Sigma K)_n$, and the augmentation $0: \mathbb{Z} \to \mathbb{Z}$. 

The \emph{join} of $(K, K^*)$ and $(L, L^*)$ is the unique augmented directed complex satisfying $\Sigma((K, K^*) \join (L, L^*)) \simeq \Sigma(K,K^*)\otimes \Sigma(L,L^*)$.

The tensor product determines a monoidal structure on $\adc$, whose unit is the augmented directed complex $I$ with $I_0 :=\mathbb{Z}$, $I_0^* := \mathbb{N}$, $I_n := 0$ for $n > 0$, and the identity on $\mathbb{Z}$ as augmentation. The join also determines a monoidal structure on $\adc$, whose unit is the augmented directed complex $0$ equal to 0 in every degree.
\end{dfn}

\begin{prop} \label{prop:adcmonoidal}
The functor $K$ is monoidal from $(\globpos, \tensor, 1)$ to $(\adc, \otimes, I)$ and from $(\globpos, \join, \emptyset)$ to $(\adc, \join, 0)$.
\end{prop}
\begin{proof}
A simple comparison of the definitions.
\end{proof}

\begin{remark}
Lax Gray products and joins do not preserve the class of positive opetopes, nor the class of simple constructible molecules. For the first class, we can take the 2-cube $K^2 = \vec{I} \tensor \vec{I}$ or the 2-simplex $\Delta^2 = \vec{I} \join 1$ as counterexamples, since both $\vec{I}$ and $1$ are positive opetopes; note however that the dual $\oppall{(\Delta^2)}$, defined later in this section, is a positive opetope.

For the second class, the 2-cube $K^2$ is a simple constructible molecule, but neither the 3-cube $K^3 = K^2 \tensor \vec{I}$, nor the join $K^2 \join 1$ are simple:
\begin{equation*}
\input{img/s2_simplejoin.tex}
\end{equation*}
\begin{equation*}
\input{img/s2_simplejoin2.tex}
\end{equation*}
\end{remark}

\begin{cons} \label{cons:jdual}
Let $P$ be an oriented graded poset, and $J \subseteq \mathbb{N}^+ = \mathbb{N}\setminus\{0\}$. Then $\oppn{J}{P}$, the \emph{$J$\nbd dual} of $P$, is the oriented graded poset with the same underlying poset as $P$, and the orientation $o'$ defined by 
\begin{equation*}
o'(c_{x,y}) := \begin{cases}
		-o(c_{x,y}), & \dmn{x} \in J, \\
		o(c_{x,y}), & \dmn{x} \not\in J,
	\end{cases}
\end{equation*}
for all elements $x, y \in P$ such that $x$ covers $y$. This extends to an involutive endofunctor $\oppn{J}{-}$ on $\ogpos$.
\end{cons}

\begin{prop} \label{prop:oppglobpos}
Let $P$ be a constructible directed complex, $J \subseteq \mathbb{N}^+$. Then $\oppn{J}{P}$ is a constructible directed complex.
\end{prop}
\begin{proof}
The underlying poset of $\oppn{J}{P}$ is the same as the underlying poset of $P$, so it is thin when $P$ is. Moreover, suppose diagram (\ref{eq:signed}) represents an interval of length 2 in $P_\bot$. In $\oppn{J}{P}_\bot$, the orientation $\alpha_1$ is flipped if and only if $\alpha_2$ is, and the same holds of the orientations $\beta_1$ and $\beta_2$. In any case, the product $\alpha_1\beta_1$ is flipped if and only if the product $\alpha_2\beta_2$ is.

Next, we show that for any constructible molecule $U$ in $P$, the corresponding subset $\oppn{J}{U}$ of $\oppn{J}{P}$ is a constructible molecule, and if $V \sqsubseteq U$, then $\oppn{J}{V} \sqsubseteq \oppn{J}{U}$. If $U$ is a 0-globe, this is obvious. Suppose $U$ is a constructible $n$\nbd molecule; then $\bord{}{\alpha}\oppn{J}{U} = \oppn{J}{\bord{}{\pm\alpha}U}$, depending on whether $n \in J$ or not, and in any case a constructible $(n-1)$\nbd molecule by the inductive hypothesis. If $U$ is atomic, this is sufficient. Otherwise, $U$ splits as $U_1 \cup U_2$. It is straightforward to verify that, if $n \in J$, then $\oppn{J}{U}$ splits as $\oppn{J}{U_2} \cup \oppn{J}{U_1}$, and if $n \notin J$, then $\oppn{J}{U}$ splits as $\oppn{J}{U_1} \cup \oppn{J}{U_2}$.

Since the closure of an element $x$ in $\oppn{J}{P}$ is dual to its closure in $P$, this proves that $\oppn{J}{P}$ is a constructible directed complex.
\end{proof}

There are three particularly interesting instances of $J$\nbd duals:
\begin{enumerate}
	\item $\oppall{P} := \oppn{\mathbb{N}^+}{P}$,
	\item $\opp{P} := \oppn{J_\text{odd}}{P}$ where $J_\text{odd} = \{2n - 1 \,|\, n > 0\}$, and
	\item $\coo{P} := \oppn{J_\text{even}}{P}$ where $J_\text{even} = \{2n \,|\, n > 0\}$.
\end{enumerate}
Clearly, $\oppall{P} = \opp{(\coo{P})} = \coo{(\opp{P})}$. 

\begin{prop} \label{prop:opglob}
Let $P, Q$ be two oriented graded posets. Then:
\begin{enumerate}[label=(\alph*)]
	\item $x \tensor y \mapsto y \tensor x$ defines an isomorphism between $\opp{(P \tensor Q)}$ and $\opp{Q} \tensor \opp{P}$ and between $\coo{(P \tensor Q)}$ and $\coo{Q} \tensor \coo{P}$;
	\item $x \join y \mapsto y \join x$ defines an isomorphism between $\opp{(P \join Q)}$ and $\opp{Q} \join \opp{P}$.
\end{enumerate}
Consequently, $x \tensor y \mapsto x \tensor y$ defines an isomorphism between $\oppall{(P \tensor Q)}$ and $\oppall{P} \tensor \oppall{Q}$.
\end{prop}
\begin{proof}
Let $x \tensor y \in \opp{(P \tensor Q)}$, with $\dmn{x} = n$ and $\dmn{y} = m$. Then $x \tensor y$ covers $x' \tensor y$ in $\opp{(P \tensor Q)}$ with orientation $\alpha$ if and only if $x \tensor y$ covers $x' \tensor y$ in $P \tensor Q$ with orientation $(-)^{n+m}\alpha$, if and only $x$ covers $x'$ in $P$ with orientation $(-)^{n+m}\alpha$, if and only if $x$ covers $x'$ in $\opp{P}$ with orientation $(-)^n(-)^{n+m}\alpha = (-)^m\alpha$, if and only if $y \tensor x$ covers $y \tensor x'$ in $\opp{Q} \tensor \opp{P}$ with orientation $\alpha$. 

Similarly, $x \tensor y$ covers $x \tensor y'$ in $\opp{(P \tensor Q)}$ with orientation $\alpha$ if and only if $x \tensor y$ covers $x \tensor y'$ in $P \tensor Q$ with orientation $(-)^{n+m}\alpha$, if and only if $y$ covers $y'$ in $Q$ with orientation $(-)^n(-)^{n+m}\alpha = (-)^m\alpha$, if and only if $y$ covers $y'$ in $\opp{Q}$ with orientation $(-)^m(-)^m\alpha = \alpha$, if and only if $y \tensor x$ covers $y' \tensor x$ in $\opp{Q} \tensor \opp{P}$ with orientation $\alpha$.

The proofs for $\coo{(P \tensor Q)}$ and $\opp{(P \join Q)}$ are analogous.
\end{proof}

\begin{remark}
There is no isomorphism between $\coo{(P \join Q)}$ and $\coo{Q} \join \coo{P}$, unless $P$ or $Q$ is empty: if $x \in P$ and $y \in Q$ are 0-dimensional, then $x \join y$ covers $y$ in $\coo{(P \join Q)}$ with orientation $+$, but $y \join x$ covers $y$ in $\coo{Q} \join \coo{P}$ with orientation $-$.
\end{remark}

Next, we focus on the extension of these operations to constructible polygraphs.

\begin{dfn}
Let $\globe_+$ be the full subcategory of $\globpos$ containing $\globe$ and the empty oriented graded poset $\emptyset$. An \emph{augmented constructible polygraph} is a presheaf on $\globe_+$. Augmented constructible polygraphs and their morphisms form a category $\cpol_+$.

For any constructible polygraph $X$, the \emph{trivial augmentation} $X_+$ of $X$ is the augmented constructible polygraph extending $X$ by $X(\emptyset) = \{*\}$. This determines a full and faithful functor $(-)_+: \cpol \to \cpol_+$ in the obvious way, making $\cpol$ a reflective subcategory of $\cpol_+$.
\end{dfn}

\begin{cons}
By Proposition \ref{thm:globpostensor}, the lax Gray product restricts to a monoidal structure on $\globe$, and by Proposition \ref{lem:globjoin}, the join restricts to a monoidal structure on $\globe_+$. There is a canonical way of extending a monoidal structure on a small category to a monoidal biclosed structure on its category of presheaves, by Day convolution \cite{day1970closed}. 

For the lax Gray product, the monoidal product of two constructible polygraphs $X, Y$ is defined by the coend
\begin{equation*}
	X \tensor Y := \int^{U, V \in \globe} X(U) \times Y(V) \times \homset{\globe}(-,U \tensor V),
\end{equation*}
the left hom $\homoplax{X}{Y}$ by the end
\begin{equation*}
	\homoplax{X}{Y} := \int_{U \in \globe} \homset{\cat{Set}}(X(U), Y(- \tensor U)),
\end{equation*}
and the right hom $\homlax{X}{Y}$ by the end
\begin{equation*}
	\homlax{X}{Y} := \int_{U \in \globe} \homset{\cat{Set}}(X(U), Y(U \tensor -)).
\end{equation*}
The unit is the Yoneda embedding of $1$.

Similarly, for the join, given two augmented constructible polygraphs $X,Y$, we have
\begin{align*}
	X \jointil Y & := \int^{U, V \in \globe_+} X(U) \times Y(V) \times \homset{\globe_+}(-,U \join V), \\
	\joinl{X}{Y} & := \int_{U \in \globe_+} \homset{\cat{Set}}(X(U), Y(- \join U)), \\
	\joinr{X}{Y} & := \int_{U \in \globe_+} \homset{\cat{Set}}(X(U), Y(U \join -)).
\end{align*}
This defines a monoidal biclosed structure on $\cpol_+$, whose unit is the Yoneda embedding of $\emptyset$. This does \emph{not} reflect onto a monoidal biclosed structure on $\cpol$, because $\cpol$ is not an exponential ideal of $\cpol_+$ in the biclosed sense. On the other hand, $\cpol$ is closed under joins in $\cpol_+$, because if $X(\emptyset) = Y(\emptyset) = \{*\}$, then
\begin{equation*}
	X \jointil Y(\emptyset) = X(\emptyset) \times Y(\emptyset) \times \homset{\globe_+}(\emptyset,\emptyset) \simeq \{*\}.
\end{equation*}
Thus if $r: \cpol_+ \to \cpol$ is the restriction functor, we can define $X \join Y := r(X_+ \jointil Y_+)$ for two constructible polygraphs, and obtain a monoidal structure on $\cpol$.
\end{cons}

\begin{remark}
By restriction to full subcategories of $\globe$ and $\globe_+$, we obtain the category of pre-cubical sets with the lax Gray product, and the category of semi-simplicial sets with the join, as full monoidal subcategories of $(\cpol, \tensor, 1)$ and of $(\cpol, \join, \emptyset)$, respectively.
\end{remark}

\begin{remark}
Because $\emptyset$ is initial in $\cpol$, we have natural inclusions
\begin{equation*}
	\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (-2,0) {$X$};
	\node[scale=1.25] (1) at (-.5,0) {$X \join \emptyset$};
	\node[scale=1.25] (2) at (2,0) {$X \join Y,$};
	\draw[1cinc] (0) to node[auto] {$\sim$} (1);
	\draw[1cinc] (1) to node[auto] {$\mathrm{id}_X \join !$} (2);
\end{tikzpicture}
\end{equation*}
\begin{equation*}
	\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (-2,0) {$Y$};
	\node[scale=1.25] (1) at (-.5,0) {$\emptyset \join Y$};
	\node[scale=1.25] (2) at (2,0) {$X \join Y,$};
	\draw[1cinc] (0) to node[auto] {$\sim$} (1);
	\draw[1cinc] (1) to node[auto] {$! \join \mathrm{id}_Y$} (2);
\end{tikzpicture}
\end{equation*}
for all $X$ and $Y$.
\end{remark}


\begin{cons}
Let $X$ be a constructible polygraph, and $J \subseteq \mathbb{N}^+$. The \emph{$J$\nbd dual} $\oppn{J}{X}$ of $X$ is the constructible polygraph defined by $\oppn{J}{X}(-) := X(\oppn{J}{-})$. For each $J$, this defines an endofunctor $\oppn{J}{-}$ on $\cpol$.

In particular, we write $\oppall{X}$, $\opp{X}$, and $\coo{X}$ for $X(\oppall{(-)})$, $X(\opp{(-)})$, and $X(\coo{(-)})$, respectively.
\end{cons}

\begin{prop}
There exist canonical isomorphisms of constructible polygraphs
\begin{equation*}
	\opp{(X \tensor Y)} \simeq \opp{Y} \tensor \opp{X}, \quad \quad 	\coo{(X \tensor Y)} \simeq \coo{Y} \tensor \coo{X}, \quad \quad \oppall{(X \tensor Y)} \simeq \oppall{X} \tensor \oppall{Y},
\end{equation*}
\begin{equation*}
	\opp{(X \join Y)} \simeq \opp{Y} \join \opp{X},
\end{equation*}
natural in $X$ and $Y$.
\end{prop}
\begin{proof}
We have
\begin{align*}
	\opp{(X \tensor Y)} = & \int^{U, V \in \globe} X(U) \times Y(V) \times \mathrm{Hom}_\globe(\opp{(-)},U \tensor V) = \\
	= & \int^{U, V \in \globe} X(\opp{U}) \times Y(\opp{V}) \times \mathrm{Hom}_\globe(\opp{(-)},\opp{U} \tensor \opp{V})
\end{align*}
because the coend is over all atoms, and any atom $U$ is equal to $\opp{(\opp{U})}$. By Proposition \ref{prop:opglob}, this is isomorphic to 
\begin{align*}
	& \int^{U, V \in \globe} X(\opp{U}) \times Y(\opp{V}) \times \mathrm{Hom}_\globe(\opp{(-)},\opp{(V \tensor U)}) \simeq \\
	\simeq & \int^{U, V \in \globe} \opp{X}(U) \times \opp{Y}(V) \times \mathrm{Hom}_\globe(-,V \tensor U) = \opp{Y} \tensor \opp{X}.
\end{align*}
The other natural isomorphisms are constructed in the same way, using the appropriate parts of Proposition \ref{prop:opglob}.
\end{proof}

\begin{cons} \label{cons:slices}
Like the category of $\omega$\nbd categories with the join defined by Ara and Maltsiniotis \cite{ara2016joint}, $\cpol$ with the join is not biclosed, but it is locally biclosed, in the following sense. For all constructible polygraphs $X, Y$, we have an augmented constructible polygraph $\joinr{X_+}{Y_+}$, and
\begin{align*} 
	\joinr{X_+}{Y_+}(\emptyset) & \simeq \homset{\cpol_+}(\emptyset, \joinr{X_+}{Y_+}) \simeq \\
		& \simeq \homset{\cpol_+}(X_+\jointil\emptyset, Y_+) \simeq \homset{\cpol_+}(X_+, Y_+),
\end{align*}
which is naturally isomorphic to $\homset{\cpol}(X, Y)$ since $\cpol$ is a full subcategory of $\cpol_+$.

For each atom $U$, let $!: \emptyset \incl U$ be the unique inclusion of $\emptyset$ into $U$. For each morphism $f: X \to Y$ of constructible polygraphs, we define a sub-presheaf $\slice{f}{Y}$ of $\joinr{X_+}{Y_+}$ by
\begin{equation*}
	\slice{f}{Y}(U) := \{x \in \joinr{X_+}{Y_+}(U) \,|\, !^*x = f\},
\end{equation*}
where the identity is interpreted via $\joinr{X_+}{Y_+}(\emptyset) \simeq \homset{\cpol}(X, Y)$.

This is well-defined as a sub-presheaf, and $\slice{f}{Y}(\emptyset) = \{f\}$, so we can identify it with a constructible polygraph.

\begin{remark} \label{rmk:slice_diagram}
Via the natural isomorphisms
\begin{equation*}
	\joinr{X_+}{Y_+}(U) \simeq \homset{\cpol_+}(U, \joinr{X_+}{Y_+}) \simeq \homset{\cpol}(X \join U, Y),
\end{equation*}
the elements of $\slice{f}{Y}(U)$ correspond to morphisms $h: X\join U \to Y$ such that 
\begin{equation} \label{eq:slice_diagram}
	\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (-1.25,-1.25) {$X \join U$};
	\node[scale=1.25] (1) at (0,0) {$X$};
	\node[scale=1.25] (2) at (1.25,-1.25) {$Y$};
	\draw[1cincl] (1) to (0);
	\draw[1c] (1) to node[auto] {$f$} (2);
	\draw[1c] (0) to node[auto] {$h$} (2);
\end{tikzpicture}
\end{equation}
commutes.
\end{remark}

Similarly, there is a sub-presheaf $\sliceco{Y}{f}$ of $\joinl{X_+}{Y_+}$, defined by
\begin{equation*}
	\sliceco{Y}{f}(U) := \{x \in \joinl{X_+}{Y_+}(U) \,|\, !^*x = f\}.
\end{equation*}
\end{cons}

\begin{prop} \label{prop:locallybiclosed}
The assignments $(f: X \to Y) \mapsto \slice{f}{Y}$ and $\sliceco{Y}{f}$ of Construction \ref{cons:slices} extend to functors $\slice{X}{\cpol} \to \cpol$. These functors are right adjoints to the functors $\cpol \to \slice{X}{\cpol}$ defined, respectively, by
\begin{equation} \label{eq:join_first_incl}
	f: Y \to Z \quad \quad \mapsto \quad \quad 
	\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (-1.25,-1.25) {$X \join Y$};
	\node[scale=1.25] (1) at (0,0) {$X$};
	\node[scale=1.25] (2) at (1.25,-1.25) {$X \join Z$};
	\draw[1cincl] (1) to (0);
	\draw[1cinc] (1) to (2);
	\draw[1c] (0) to node[auto] {$\mathrm{id}_X \join f$} (2);
	\node[scale=1.25] at (2,-1.4) {,};
\end{tikzpicture}
\end{equation}
\begin{equation*} 
	f: Y \to Z \quad \quad \mapsto \quad \quad 
	\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (-1.25,-1.25) {$Y \join X$};
	\node[scale=1.25] (1) at (0,0) {$X$};
	\node[scale=1.25] (2) at (1.25,-1.25) {$Z \join X$};
	\draw[1cincl] (1) to (0);
	\draw[1cinc] (1) to (2);
	\draw[1c] (0) to node[auto] {$f \join \mathrm{id}_X$} (2);
	\node[scale=1.25] at (2,-1.4) {.};
\end{tikzpicture}
\end{equation*}
\end{prop}
\begin{proof}
Given a morphism of constructible polygraphs $g: Y \to Z$, the morphisms 
\begin{equation*}
	\joinr{X_+}{g}: \joinr{X_+}{Y_+} \to \joinr{X_+}{Z_+}, \quad \quad \quad \joinl{X_+}{g}: \joinl{X_+}{Y_+} \to \joinl{X_+}{Z_+}
\end{equation*}
restrict to morphisms 
\begin{equation*}
	g_*: \slice{f}{Y} \to \slice{(f;g)}{Z}, \quad \quad \quad g_*: \sliceco{Y}{f} \to \sliceco{Z}{(f;g)}.
\end{equation*}
This is clear from the reformulation of Remark \ref{rmk:slice_diagram}: if $(\ref{eq:slice_diagram})$ commutes, so does the post-composition of the triangle with $g$. This gives functoriality of $\slice{f}{Y}$.

The adjunction with $(\ref{eq:join_first_incl})$ is also almost obvious from Remark \ref{rmk:slice_diagram}: a morphism $(X \incl X \join Y) \to (h: X \to Z)$ is a commutative triangle
\begin{equation*}
	\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (-1.25,-1.25) {$X \join Y$};
	\node[scale=1.25] (1) at (0,0) {$X$};
	\node[scale=1.25] (2) at (1.25,-1.25) {$Z$};
	\draw[1cincl] (1) to (0);
	\draw[1c] (1) to node[auto] {$h$} (2);
	\draw[1c] (0) to node[auto] {$k$} (2);
	\node[scale=1.25] at (1.75,-1.4) {,};
\end{tikzpicture}
\end{equation*}
which via the join-right hom adjunction corresponds to a morphism $\widehat{k}: Y_+ \to \joinr{X_+}{Z_+}$ with the property that $\widehat{k}(y) \in \slice{f}{Y}$ for all $y \in Y$.

The case of $\sliceco{Y}{f}$ is completely analogous.
\end{proof}

In \cite{ara2016joint}, the analogue of $\slice{f}{Y}$ for $\omega$\nbd categories is called the \emph{generalised slice} of $Y$ under $f$. Ara and Maltsiniotis reserve the name of generalised slice of $Y$ over $f$ for $\oppall{(\slice{\oppall{f}}{\oppall{Y}})}$, rather than $\sliceco{Y}{f}$.
{"config": "arxiv", "file": "1806.10353/monoidal.tex"}