TITLE: (Certain) colimit and product in category of topological spaces
QUESTION [7 upvotes]: Consider the diagram 
$$(*):\;\;\;X_0 \stackrel{i_0}\hookrightarrow X_1 \stackrel{i_1}\hookrightarrow  X_2 \stackrel{i_2}\hookrightarrow \cdots $$
in category of topological spaces. Denote $I$ the unit interval $[0,1]$ with the standard topology.
Then we also have
$$(*)\times I:\;\;\;X_0\times I \stackrel{i_0\times 1_I}\hookrightarrow X_1 \times I \stackrel{i_1\times 1_I}\hookrightarrow  X_2 \times I \hookrightarrow \cdots $$
and colimits of those two diagrams, which are basically
$$\mathrm{colim}(*)=\left(\coprod_{n =0}^\infty X_n\right)/{\sim}, \;\;\;  x \sim i_j(x), \; j \in \mathbb{N}, \; x \in X_j,$$
$$\mathrm{colim}((*)\times I)=\left(\coprod_{n =0}^\infty X_n \times I\right)/{\sim}, \;\;\; (x,t) \sim (i_j(x),t), \; j \in \mathbb{N}, \; x \in X_j, t \in I.$$
(Question 0: is this correct?)
I am interested in the map $$f:(\operatorname{colim}(*))\times I \rightarrow \mathrm{colim}((*)\times I)$$
defined as $([x],t)\mapsto [(x,t)]$, where $[-]$ denotes taking the equivalence classes with respect to the considered equivalences. Obviously it is correctly defined as a map.
Question 1: Is this map continous? That is, are the two objects homeomorphic? (I believe one can obtain the inverse map $f^{-1}$ using the universal property of $\mathrm{colim}((*)\times I)$, hence $f^{-1}$ is continuous).
I have more information about the maps $i_j$. For example, they are closed maps.
This is a part of a homework assignment, so I would appreciate hints instead of full answer.
(Maybe some context and motivation: My goal is to define a homotopy $F:\mathrm{colim}(*))\times I \rightarrow \mathrm{colim}(*)$ using easy-to-describe homotopies $F_n:X_n\times I \rightarrow X_{n+1}$, but these induce "only" a map $(\operatorname{colim }F_n): \operatorname{colim}((*)\times I) \rightarrow \operatorname{colim}(*)$, so I am trying to change the domain somehow - so ultimately, I need only the composite map $(\operatorname{colim }F_n) \circ f$ to be continuous.)

REPLY [0 votes]: There is a very nice proof in terms of adjoint functors. We want to test the conjectured isomorphism against all topological spaces, that is
$$ \textrm{Hom}_{Top} \left ( \varinjlim_{d \in D } X(d) \times I , Y \right ) \stackrel{?}{\simeq} \textrm{Hom}_{Top} \left ( \left (\varinjlim_{d \in D } X(n) \right ) \times I , Y \right )$$
Here $D$ is whatever category, $X: D \to \textrm{Top}$ is whatever diagram of spaces, and $Y$ is any space.
Our strategy is to use adjoint functors and bring everything on the right side of the hom, then check if we actually have an isomorphism. This could seem very clever the first time, but it's sort of the unique passage you can do to do some manipulations.
First thing we have to be able to "bring on the other side" colimits: this comes from the adjunction
$$ \textrm{Hom}_{Top} \left ( \varinjlim_{d \in D } F(d) , Z \right ) \simeq \textrm{Nat}_{D} ( F, \Delta(Z))$$
for any functor $F: D \to \textrm{Top}$ and space $Z$, where $\Delta: \textrm{Top} \to \textrm{Top}^{\mathbb{N}} $ is the constant functor.
Second, we have to be able to bring on the other side the $_ \times I$ operation. This is possible since $I$ is locally compact and the exponential law holds:
$$ \textrm{Hom}_{\textrm{Top}}(A \times I, B) \simeq \textrm{Hom}_{\textrm{Top}}(A , B^I) $$
The general theorem is not easy, but in this special case (that is, the space being the interval) you should be able to prove it yourself.
Now let's put everything together: we have
$$\textrm{Hom}_{Top} \left ( \varinjlim_{d \in D } X(d) \times I , Y \right )  \simeq \textrm{Nat}_{D} ( X \times I , \Delta(Y)) \simeq \textrm{Nat}_{D} ( X  , \Delta(Y)^I) $$
Now note that taking maps from $I$ to $\Delta(Y)$ componentwise is the same of taking the constant functor $\Delta(Y^{I})$, since $\Delta(Y)$ is a constant functor!! Thus we can continue, using adjunctions in the other way around:
$$ \simeq \textrm{Nat}_{D} ( X  , \Delta(Y^I)) \simeq \textrm{Hom}_{Top} \left ( \varinjlim_{d \in D } X(d)  , Y^I \right )  \simeq \textrm{Hom}_{Top} \left ( \left ( \varinjlim_{d \in D } X(d) \right ) \times I , Y \right ) $$
Wow, done! As you can see, the (hard) commutation of colim and $\times I$ becomes the (easy) commutation of $\Delta$ and $(-)^I$ in the adjoint point of view.
{"set_name": "stack_exchange", "score": 7, "question_id": 833042}