Conventions for definitions of the cap product In singular (co)homology, if $\alpha\in C^*(X)$ and $x\in C_*(X)$, then the cap product $\alpha \cap x$ is generally defined by the following process:


*

*Apply to $x$ the diagonal map $C_*(X)\to C_*(X\times X)$ followed by some choice of Alexander-Whitney chain equivalence $ C_*(X\times X)\to C_*(X)\otimes C_*(X)$ to obtain an element $\sum y_i\otimes z_i$. 

*Apply $\alpha$ to $\sum y_i\otimes z_i$ by the slant product, or, in other words and roughly speaking, apply $\alpha$ to ``half of the factors''. 

*Depending on your favorite conventions (and what you're trying to accomplish), there may also be a sign (which might also be part of your definition of the slant product - but let's ignore this).
The reason I'm being so cagey with wording in part 2 is directly related to my question: In almost all major textbook sources I have consulted, step 2 is performed by forming $\sum y_i \alpha(z_i)$, which strikes me as somewhat unnatural, forcing the $\alpha$, which starts off on the left to jump all the way over the $y_i$ terms to get to the $z_i$ terms on the right. Is there a good mathematical reason for this convention? Why not define the cap product to be $\sum \alpha(y_i) z_i$?
The one major exception to this convention seems to be Hatcher. He does form $\sum \alpha(y_i) z_i$, but he also writes cap products as $x\cap \alpha$, so his cochain also has to jump, but it jumps over the $z$s instead!
(For the record, I'm not asking this question out of idle pickiness. Jim McClure and I have been doing a lot of work with cap products recently, and we're trying to be consistent amongst various conventions for various issues, but preferably with good reasons thrown in!) 
 A: This is just an expanded version of Tyler's comment, I think.
Let's use a, b, c for cochains, x, y, z for chains, [a,x] for the value of a cochain on a chain. I'll be lazy and write $ab$ for $a\cup b$ and $ax$ for $a\cap x$. Let $[a\otimes b,y\otimes z]=(-1)^{|b||y|}[a,y][b,z]$.
I like to define $bx$ in such a way that $[a,bx]$ is $[ab,x]$. Then associativity and unit of cup makes cap into a module structure: $(bc)x=b(cx)$ because $[a,(bc)x]=[a(bc),x]=[(ab)c,x]=[ab,cx]=[a,b(cx)]$, and $1x=x$ follows from $[a,1x]=[a1,x]=[a,x]$. 
If you have a product $a\otimes b\mapsto ab$ of cochains and a coproduct of chains defined in such a way that $[ab,x]=[a\otimes b,y_i\otimes z_i]$ where the coproduct of $x$ is $\Sigma y_i\otimes z_i$, then that means I have to define $bx$ to be $\Sigma (-1)^{|y_i||b|}[b,z_i]y_i$, so as to get
$[a,bx]=[ab,x]=[a\otimes b,\Sigma y_i\otimes z_i]=\Sigma (-1)^{|y_i||b|}[a,y_i][b,z_i]=[a,\Sigma (-1)^{|y_i||b|}[b,z_i]y_i]$.
The sign is a bit unexpected, as is the jump you mention, but it's worth it for the neat formulas in the third paragraph above.
It's not a bad idea to define $xa$ as well. I'd do it by first declaring $[x,a]$ to be $(-1)^p[a,x]$ when $a$ is a $p$-cochain and $x$ is a $p$-chain, then defining $xa$ in such a way that $[xa,b]=[x,ab]$. This insures $x(ab)=(xa)b$ and $x1=x$. The chain-level formula is no better and no worse than the formula for $ax$.
Of course, $ax$ and $xa$ end up differing only by a sign when you get to homology, but the sign is hard to remember; in working it out you have to use the commutativity law for the cup product.
