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In basic algebraic topology, we know the following well-known chain homotopy theorem:

Let $X$ be a topological space and $I=[0,1]$ be the unit interval. Let $S_*(X)$ and $S_*(X\times I)$ be the singular chain groups of $X$ and $X\times I$ respectively. Let $\tau_0$ and $\tau_1$ be the two natural inclusions $X\hookrightarrow X\times I$. Then $\tau_0$ and $\tau_1$ induce homotopic chain maps on the singular chains. In more details, there exists a map $$P:S_*(X)\rightarrow S_{*+1}(X\times I)$$ such that $$ \partial P+P\partial=(\tau_0)_*-(\tau_1)_*. $$

We can talk a little about $P$. Let $\Delta^n$ denote the $n$-dimensional standard simplex, we can decompose $\Delta^n\times I$ into union of $n+1$-dimensional simplices. Let $a_0,a_1,\ldots, a_n$ denote the vertices of $\Delta^n$ and $b_0,b_1$ denote the two vertices of $I$, then the vertices of $\Delta^n\times I$ can be denoted as a pair $(a_i,b_j)$, $0\leq i\leq n, j=0,1$. Now we obtain the decompostion $$ \Delta^n\times I=\bigcup_i [(a_0,b_0)\ldots(a_i,b_0)(a_i,b_1)\ldots(a_n,b_1)] $$

In fact the map $P$ is defined using this decomposition as follows. Let $\sigma\in S_n(X)$ be a $n$-dimensional singular chain in $X$, i.e. $\sigma$ is a map $\Delta^n\rightarrow X$. then $\sigma\times \text{id}$ is a map $\Delta^n\times I\rightarrow X\times I$ and we define $P(\sigma)\in S_{n+1}(X\times I)$ to be $$ P(\sigma)([e_0,\ldots,e_{n+1}]):=\sum_{i=0}^{n}(-1)^i(\sigma\times \text{id})\circ [(a_0,b_0)\ldots(a_i,b_0)(a_i,b_1)\ldots(a_n,b_1)] $$

It is not difficult to check that the $P$ defined above stisfies $\partial P(\sigma)+P\partial \sigma=(\tau_0)_*\sigma-(\tau_1)_*\sigma$. For more details see Hatcher http://www.math.cornell.edu/~hatcher/AT/ATch2.pdf page 111-112. Our notation is slightly different from his.

We notice that the interval $I$ can be identified with the standard $1$-simplex $\Delta^1$. In this viewpoint $\tau_0$ and $\tau_1$ can be considered as the two "coface" maps in the cosimplicial set.

Now it is natural to consider the higher dimensional generalizations for the above result: for any $m,n \geq 0$, we can also decompose $\Delta^n\times \Delta^m$ as unions of $n+m$-dimensional simplices: The vertices of $\Delta^n\times \Delta^m$ can be denoted by $(a_i,b_j)$, $0\leq i\leq n, 0\leq j\leq m$.

With these notation, we can decompose $\Delta^n\times \Delta^m$ into unions of $n+m$-dimensional simplices: the adjacent vertices of the $(n+m)$ vertices must have the form $$ (a_i,b_j)(a_{i+1},b_j) \text{ or } (a_i,b_j)(a_i,b_{j+1}). $$ It is illustrative to consider a $n\times m$ lattice and we want to move form $(0,0)$ to $(n,m)$ in $n+m$ steps, and each step we can only move to the right or downwards.

$\textbf{My question}$ is: do we have an expression of higher chain homotopies involves the coface maps between $S_*(X)$, $S_*(X\times \Delta^1)$, $\ldots$, $S_*(X\times\Delta^m)$?

Maybe the construction already exists and is well-known to experts and any references are really appreciated.

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Why don't you apply Eilenberg-Zilber? Then the problem reduces to writing down an explicit contraction of $C_{\bullet}(\Delta^m)$ which IIRC is straightforward. –  Qiaochu Yuan Dec 14 '13 at 10:03
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There is also a lot to be said for doing singular homology cubically. Massey's book on "Singular homology theory" (Springer, 1980) takes this approach. One advantage of cubes is the rule $I^m \times I^n \cong I^{m+n}$. –  Ronnie Brown Dec 14 '13 at 10:57
    
@QiaochuYuan May I ask what does "IIRC" stand for? –  Zhaoting Wei Dec 14 '13 at 15:52
    
lmgtfy.com/?q=iirc –  Qiaochu Yuan Dec 14 '13 at 19:00
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@QiaochuYuan, I think that link was overdoing it. It might be best to simply avoid using "internet acronyms" here. –  Scott Morrison Dec 14 '13 at 23:10

1 Answer 1

The key idea to look at is that of a shuffle. These come into the Eilenberg -Zilber theorem, but as they enable the combinatorics of the $m+n$ simplices of $\Delta^n\times \Delta^m$ to be given the various fairly elementary moves across higher simplices to be given. Really, however, it seems to me, that the additive chain complex groups are not the place to study this as the singular complex given as a simplicial set enables the full structure to be exhibited explicitly. There is a bit more combinatorial listing to be done but it is worth it.

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Thank you very much for your answer and would you like to explain a little bit about "the additive chain complex groups are not the place to study this as the singular complex given as a simplicial set enables the full structure to be exhibited explicitly."? Are additive chain complex and singular complex the same thing? –  Zhaoting Wei Dec 14 '13 at 17:52
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The singular chain groups that you mention are all free Abelian groups on a simplicial set of singular simplices in $X$. Look at one of the fairly elementary introductions to simplicial set theory (I like the old paper by Curtis:Simplicial Homotopy Theory, Advances in Math., 6,(1971),107 – 209, but there are many other introductions).The structure of product simplicial sets is very simple, so at least to start with, homotopies etc. are quite nice to write down.To understand what higher homotopies might be you can work with the simplicial function spaces and then specialise down to the prisms. –  Tim Porter Dec 14 '13 at 18:39
    
The combinatorial version of homotopies is then easy to give. Trying to give the higher homotopies simplicially is a case of studying the shuffles .... and they are fun! –  Tim Porter Dec 14 '13 at 18:41

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