TITLE: Singular complex is a delta complex
QUESTION [4 upvotes]: If I understand correctly, $\Delta$-complex on a space $X$ is defined to be a collection $\Delta(X)$ of cont. funtions $\sigma:\Delta^n\to X$ such that:
1) restriction of $\sigma$ to any face is in $\Delta(X)$
2) restriction of $\sigma$ to the interior is injective
3) $A \subseteq X$ is open if every $\sigma^{-1}(A)$ is open  
Can anyone explain me how a singular complex $S(X)$ can be regarded as a $\Delta$-complex considering the condition (2)? I have read some books but I still don't get it.
Thank you for your help.

REPLY [4 votes]: The singular complex $S(X)$ is not $Δ$-complex structure on space $X$, but it is a completely new space constructed as a geometric realization of an abstract $Δ$-structure given by singular simplices of original space $X$.
Example: Take $X = \{*\}$ the one-point space. For each $n$ there is only one singular $n$-simplex – the constant one. The singular complex $Δ(X)$ looks as follows. It has one $n$-cell for each $n$. Each of them represents the corresponding $n$-simplex. So the $0$-skeleton of $Δ(X)$ is just a point. In $1$-skeleton, we glue a line segment to the point with both endings. In $2$-skeleton, we glue a border of a triangle to the line segment. Each segment of the border is glued to the only line in $1$-skeleton. The first one is glued forwards, the second one backwards and the third one forwards. We continue in this manner with all $n$-cells. So $Δ(X)$ of the one-point space is actually infinite-dimensional $Δ$-complex with one cell in each dimension.
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