I am having trouble to understand how the image of an element $a\otimes b\in C_p(X)\otimes C_q(Y)$ under the Eilenberg-Zilber map on singular chains is an element of $C_{p+q}(X\times Y)$.
I am using the explicit expression of the map given in nlab which works in general for any simplicial abelian group, or the one in page 7 of this paper given under the name of EML operator for normalized chains.
So, let $\Delta^n$ be the simplex of dimension $n$ and consider the degenerancy maps $s^i:\Delta^n\to\Delta^{n-1}$. Let $a:\Delta^p\to X$ and $b:\Delta^q\to Y$. We may define degenerancy maps $s_i:C_{p}(X)\to C_{p+1}(X)$ by $s_i(a)=a\circ s^i$. Now, up to sign, when I apply the Eilenberg-Zilber map to $a\otimes b$ I get
$$\sum_{(\mu, \nu)}s_\nu(a)\times s_\mu(b)$$
where $(\mu,\nu)$ are $(p,q)$–shuffles and $s_\mu$ is a composition of degenerancy maps. Let me do an example so show my problem with this definition.
Let $a:[0,1]\to X$ and $b:[0,1]\to Y$ two 1-simplices. There are two $(1,1)$-shuffles which are the 2 permutations of $S_2$. Therefore, there are only two degenerancy maps to take into account, $s_0$ and $s_1$.The Eilenberg-Zilber map here is
$$\nabla(a\otimes b)=s_1(a)\times s_0(b)-s_0(a)\times s_1(b)$$
My problem is that a chain in $C_{p+q}(X\times Y)$ should be a map $\Delta^{p+q}\to X\times Y$, which in this example means $\Delta^2\to X\times Y$. But for instance in this case, $s_1(a)=a\circ s^1: \Delta^2\to\Delta^1\to X$ and similarly $s_0(b):\Delta^2\to Y$, so what I get is a map $\Delta^2\times\Delta^2\to X\times Y$, so how exactly is this map an element of $C_2(X\times Y)$ or more generaly of $C_{p+q}(X\times Y)$?
Best Answer
Let $A$, $X$ and $Y$ be spaces. Then a map $A \to X \times Y$ is the same as the data of a map $A \to X$ and a map $A \to Y$; that is, it is uniquely determined by its components.
When you see the expression $s_\nu(a) \times s_\nu(b)$, you should read this as a map with domain $\Delta^{p+q}$ and codomain $X \times Y$, whose first coordinate is given by the map $s_\nu(a)$ and second coordinate is given by the map $s_\nu(b)$.
To be honest, this formula hides a lot of intuitive content. To me, the Eilenberg-Zilber map is: "Send $\Delta^p \otimes \Delta^q$ to the canonical triangulation of $\Delta^p \times \Delta^q$." The elements in your sum each correspond to a simplex in said triangulation.
This question was asked 5 months ago so I imagine you have now resolved whatever confusions you had, or moved on. But if you are still hoping to understand this, I strongly recommend trying to see "why" the formula above describes the standard triangulation of $\Delta^2 \times \Delta^1$. This gave me more intuition for the Eilenberg-Zilber map than anything else ever has.