Alexander Whitney map is a chain map Define $a_p : \Delta_p \longrightarrow \Delta_n$ and $b_q : \Delta_q \longrightarrow \Delta_n$ to be $\begin{cases}a_p(e_i) = e_i & p \leq n \\ b_q(e_i) = e_{n-q+i} & q \leq n \end{cases}$.
Given $\sigma \in C_n(X)$ we consider $\sigma_p^1 := \sigma \circ a_p$ and $\sigma_q^2 := \sigma \circ b_q$. Define the map $\Delta : C(X) \longrightarrow C(X) \otimes C(X)$ to be
$$\Delta \sigma_n = \sum\limits_{p+q=n} \sigma_p^1 \otimes \sigma_q^2$$
Since I didn't find any reference of this map to be a chain map I tried to prove it on my own. This is where I got stuck.
We can see that $$d \circ \Delta \sigma = \sum\limits_{p+q=n} \partial_p \sigma_p^1 \otimes \sigma_q^2 + (-1)^p \sigma_p^1 \otimes \partial_q \sigma_q^2$$
$$\Delta \circ \partial_n \sigma = \sum\limits_{p+q=n-1} (\partial_n\sigma)_p^1 \otimes (\partial_n\sigma)_q^2$$
I wrote down the boundary $\partial_n \sigma =  \sum\limits_{i =0}^n (-1)^i \sigma \circ d_i$ where $d_i$ is the map that "jumps" $i$.
I try to find some relations on this second sum, trying to find expressions for $(\sigma \circ d_i)_p^1$ and $(\sigma \circ d_i)_q^2$. I'm not sure if this is correct but what I found was
$$ \begin{cases}(\sigma \circ d_i)_p^1 = \sigma_p^1 & i > p \\ (\sigma \circ d_i)_p^1 = \sigma_{p+1}^1 \circ d_i & i \leq p \end{cases}, \hspace{0.2cm} \begin{cases}(\sigma \circ d_i)_q^2 = \sigma_q^2 & i < n-q \\ (\sigma \circ d_i)_q^2 = \sigma_{q+1}^2\circ d_{i-(n-q)+1} & i \geq n-q\end{cases}$$
In addition we have the relations $\sigma_p^1 = \sigma_{p+1}^1 \circ d_{p+1}$ and $\sigma_{q+1}^2 \circ d_0 = \sigma_q^2$.
From here I was unable to continue since splitting the two sums on the index suggested by the relations brought me nowhere, nor simplification or terms canceling out.
Any help would be appreciated, this seems the direct way to compute it but I'm not sure wheter there's an other way.
Related unanswered : Alexander Whitney: Diagonal Approximation
 A: As I said in the comments, you can really decompose your map as a composite $C(X)\to C(X\times X)\to C(X)\otimes C(X)$. The first one is induced by the diagonal $X\to X\times X$ so is clearly a chain map; and the second one is a special case of a more general AW map $C(X\times Y)\to C(X)\otimes C(Y)$.
Even worse, noting that $C(X) = C(\mathbb Z[Sing(X)])$ where $Sing(X)_n = \hom(\Delta^n, X)$ is the singular simplicial set of $X$, we see that the second map is itself a special case of the more general AW map $C(A\otimes B)\to C(A)\otimes C(B)$, where $A$ and $B$ are simplicial abelian groups (and the tensor product is pointwise, so that $(A\otimes B)_n = A_n\otimes B_n$ - you should check that $\mathbb Z[Sing(X)]\otimes \mathbb Z[Sing(Y)]\cong \mathbb Z[Sing(X\times Y)]$ as an exercise.
This morphism is described e.g. here, it's defined by $a\otimes b \mapsto \sum_{p+q=n} \tilde d^pa \otimes d^qb$, where $\tilde d$ is the "front" face $[p]\to [p+q], i\mapsto i$, and $d^q$ is the back face, $[q]\to [n], j\mapsto p+j$.
We still need to check that it's a complex morphism. Note that $\partial (\tilde d^pa \otimes d^qb) = \sum_{i=0}^p (-1)^i(d_i\tilde d^pa)\otimes d^qb + (-1)^p \sum_{j=0}^q(-1)^j\tilde d^p a\otimes (d_jd^qb)$.
Now $d_i\tilde d^p= \tilde d^{p-1}d_i$ and $d_jd^q = d^{q-1}d_{p+i}$, so this looks like
$\sum_{i=0}^p(-1)^i (\tilde d^{p-1}d_ia)\otimes d^qb + (-1)^p \sum_{j=0}^q(-1)^j\tilde d^p a\otimes (d^{q-1}d_{p+j}b)$ and so by reindexing,  $\sum_{i=0}^p (-1)^i(\tilde d^{p-1}d_ia)\otimes d^qb + \sum_{j=p}^{p+q}(-1)^j\tilde d^p a\otimes (d^{q-1}d_jb)$.
Remember you're also supposed to sum these over $p+q=n$.
We want to compare this to $\sum_{i=0}^n\sum_{p+q=n-1} (-1)^i(\tilde d^pd_ia)\otimes (d^qd_ib)$.
To do this comparison, there are two things to do :
1- Note that because you're landing in $\bigoplus_{p+q=n-1}A_p\otimes B_q$, it suffices to separate the terms and compare the terms in a single $A_p\otimes B_q$ for a single couple $(p,q)$ whose sum is $n-1$.  For the first term, that means you'll have to look at the terms coming from the couple $(p+1,q)$, or $(p,q+1)$, because these are the only ones whose differential has a nonzero coordinate in $A_p\otimes B_q$ (note that not all the differential lands in there so you have to be slightly careful)
2- You'll need to use some simplicial identities. Namely, because of how things are set up, $\tilde d^p = \tilde d^pd_i$ for a wide range of $i$'s (not all $i$'s ! it's important to figure out precisely which and think of the boundary cases), and similarly $d^q d_j = d^q$ for a wide range of $j$'s (same comment)
With these two things, you can effectively compare the two differentials and show that this is indeed a chain map. If you want more details I can add them, but let me stop here for now.
