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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.

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1 Answer 1

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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If I'm not mistaking, the point has singular simplex the final simplicial set $\Delta[0]$ whose geometric realisation is the point. I think you forgot in your description that each $n$-simplex with $n>0$ is actually degenerate. –  Pece Aug 23 '14 at 20:52
@Pece: Yes, the singular simplices are degenerate in $X$, but not in $S(X)$ by definition. We are not giving a $Δ$-structure to $X$, but constructing a $Δ$-complex, simplicial chain complex of which is singular chain complex of $X$. –  user87690 Aug 23 '14 at 21:08
Ok maybe we just have a conflict of definitions. For me, the singular simplicial complex of $X$ is the simplicial set $SX$ defined in every degree by $(SX)_n = \hom_{\mathsf{Top}}(\Delta^{\mathrm{top}}_n,X)$ where $\Delta^{\mathrm{top}}_n$ is the topological simplex of dimension $n$. The geometric realization $|-|$ is the left adjoint of $S$ : in particular, $|\Delta[n]| = \Delta^{\mathrm{top}}_n$. Then, we have $| S(\ast) | = | \Delta[0] | = \Delta^{\mathrm{top}}_0 = \ast$. –  Pece Aug 23 '14 at 21:25
Yes, we have. The singular complex I'm talking about is a certain $Δ$-complex. It is defined on page 108 in Hatcher's book so one can regard singular chain complex of a topological space as simplicial chain complex of its singular complex (the word “complex” is used in two different meanings). –  user87690 Aug 24 '14 at 8:24

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