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. 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.

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.