Alexander–Whitney map and diagonal approximation

Let $X,Y$ be spaces and $R$ a commutative unital ring, allowing $R=0$. Use ordinary, unnormalized singular chains, with the coefficient convention in Singular cochain complex with coefficients. Put $F_n(X,Y;R)={⨁}_{p+q=n}{C}_p(X;R){\otimes }_R{C}_q(Y;R),d(a\otimes b)=\mathrm{\partial }a\otimes b+(-1{)}^{p}a\otimes \mathrm{\partial }b(a\in C_p(X;R))\mathrm{.}$ Negative chain degrees and the boundary of a vertex are zero. For a simplex $\sigma :{\mathrm{\Delta }}^n\to X\times Y$, write $x={\mathrm{pr}}_X\sigma$ and $y={\mathrm{pr}}_Y\sigma$. The Alexander–Whitney map is the linear map ${\mathrm{AW}}_n:C_n(X\times Y;R)⟶F_n(X,Y;R),{\mathrm{AW}}_n(\sigma )={\sum }_{p=0}^{n}x[0,\dots ,p]\otimes y[p,\dots ,n]\mathrm{.}$ Here $x[i_0,\dots ,i_k]$ means restriction along the affine face with that ordered vertex list. The formula gives a finite chain for each generator and extends uniquely linearly; it does not require choosing representatives or fillings. Projections and the diagonal are continuous in the binary product topology. With the postcomposition maps of The induced singular chain map of a continuous map, the diagonal approximation is $D_X=\mathrm{AW}\circ {\mathrm{\Delta }}_{\mathrm{\#}}:C_{\ast }(X;R)⟶F_{\ast }(X,X;R),\mathrm{\Delta }(x)=(x,x)\mathrm{.}$

For completeness, the formula is a chain map. In $d\mathrm{AW}(\sigma )$, the first-factor faces at cut $p\ge 1$ have signs $(-1{)}^i$, $0\le i\le p$; the second-factor faces at cut $p\le n-1$ have signs $(-1{)}^j$, $p\le j\le n$. The first-factor term deleting its last vertex at cut $p$ is $(-1{)}^{p}x[0,\dots ,p-1]\otimes y[p,\dots ,n]$. It cancels the second-factor term deleting its first vertex at cut $p-1$, whose sign is $(-1{)}^{p-1}$. All remaining first-factor terms have $i<p$; they are precisely the cuts of the face of $\sigma$ omitting $i$ in which the omitted vertex lies before the cut. All remaining second-factor terms have $j>p$ and are precisely those with the omitted vertex after the cut. Each has the sign of that face in $\mathrm{\partial }\sigma$. This exhausts $\mathrm{AW}(\mathrm{\partial }\sigma )$ and proves $d\mathrm{AW}=\mathrm{AW}\mathrm{\partial }$. In degree zero both sides are zero, as required. The same face calculation shows that postcomposition commutes with $\mathrm{\partial }$; hence $D_X$ is also a chain map.

For continuous $f:X\to X^{\mathrm{\prime }}$ and $g:Y\to Y^{\mathrm{\prime }}$, face restriction commutes with postcomposition, term by term, giving $\mathrm{AW}(f\times g{)}_{\mathrm{\#}}=(f_{\mathrm{\#}}\otimes g_{\mathrm{\#}})\mathrm{AW},D_{X^{\mathrm{\prime }}}{f}_{\mathrm{\#}}=(f_{\mathrm{\#}}\otimes f_{\mathrm{\#}})D_X\mathrm{.}$ In degree zero AW sends $(x,y)$ to $x\otimes y$, preserving the augmentation that assigns $1$ to each vertex. If a factor is empty, both complexes are zero; the same holds over the zero ring. For one-point factors the formula still uses all higher singular simplices: degenerate simplices have not been discarded. The cancellation also applies to these simplices because it is an identity of face maps. Only finite sums and specified maps were used. No AC is assumed, and the arbitrary-product nonemptiness clause of the product definition is not used.