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I was given a definition and to check this is a chain complex is left as an exercise, so it is supposed to be easy. However, I am thoroughly confused by the definition.

Definition given

Let $C, D$ be chain complexes (of abelian groups, not R-modules, that one already has an answer), then the tensor product is defined to be

$$(C \otimes D)_n = \bigoplus_{p = 0}^n (C_p \otimes D_{n - p})$$

and the boundary operator is defined to be induced by the bilinear maps

$$g_{n, p}: C_p \times C_{n - p} \to (C \otimes D)_{n - 1}, \ \ \ \ (x, y) \mapsto \partial^D_p(x) \otimes y + (-1)^p x \otimes \partial_{n - p}^D(y)$$

My try at understanding it

Since $(C \otimes D)_{n - 1}$ is a direct sum, I take it to mean that we send $(x, y)$ to $n - 1$ tuple that is $0$ everywhere except on the $p - 1$'th coordinate where it is the thing specified by the map. But how does this make sense? For $p = 0$ this sends $(x, y)$ to $0 \otimes y + x \otimes 0$, but $0 \otimes y \in C_0 \otimes D_n$ which does not occur in $\bigoplus_{p = 0}^{n - 1}(C_p \otimes D_{n - 1 - p})$.

By the universal property of the tensor product, this induces maps $h_{n, p}: C_p \otimes C_{n - p} \to (C \otimes D)_{n - 1}$. Then by the universal property of the direct sum these maps induce a morphism between $(C \otimes D)_n$ and $(C \otimes D)_{n - 1}$.

Now somehow $\partial_{n - 1}^{C \otimes D} \circ \partial_{n}^{C \otimes D}$ must be equal to $0$. How does one do this with all those implicit induced maps? Call the epimorphism the tensor product $C_p \otimes D_{n - p}$ is equiped with $f$ then it is easy to prove the diagram

enter image description here

where $d$ is induced by $\partial_{n - 1}^{C \otimes D} \circ g_{n, p}$ via the universal property of the tensor product commutes, so it suffices to prove $d$ is $0$ (I think, a map from a direct product is $0$ precisely when all the components it is induced by are $0$ right?). I am at a complete loss at how to do this since $\partial_{n - 1}^{C \otimes D} \circ g_{n, p}$ is a huge mess of induced maps I do not have explicitely.

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  • $\begingroup$ The key is to not write a bunch of diagrams and induced maps but rather to just explicitly write out what the map is. $\endgroup$ Oct 3, 2019 at 21:38

1 Answer 1

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I believe you have some notation issues.

It should be $g_{n,p}(x,y) = (\partial^C_p (x) \otimes y)+ (-1)^p ( x \otimes \partial^D_{n-p}(y))$.

The first term of this lives in $C_{p-1}\otimes D_{n-p}$ and the second term lives in $C_{p} \otimes D_{n-p-1}$, so this sum should be thought of as happening in the giant direct sum $(C \otimes D)_{n-1}$ (but note that it lands in two pieces of it, rather than a single one as you described in the question).

To check that the composition is zero, it's enough to do it on homogenous pieces. The induced map thing isn't a problem: it's still exactly what you think it should be (i.e. $\partial_{n,p}(x\otimes y) = (\partial^C_p (x) \otimes y)+ (-1)^p ( x \otimes \partial^D_{n-p}(y))$ on homogeneous pieces and extend by linearity). When you do the double composition, you'll get four terms: one will die by virtue of their being a composition of $C$'s differentials, another will die by similar reasoning for $D$'s differentials, and the final two will cancel out thanks to the clever choice of sign in $g_{n,p}$.

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  • $\begingroup$ What is a homogeneous piece? $\endgroup$
    – user388557
    Oct 3, 2019 at 22:25
  • $\begingroup$ Do you mean we can write any element of $A \otimes B$ as a finite sum of elements of the form $x\otimes y$? If so, I do not understand why that is possible. $\endgroup$
    – user388557
    Oct 3, 2019 at 23:40
  • $\begingroup$ By homogenous piece I mean you need to say what it does on an element that is nonzero in exactly one direct summand of the type $C_p \otimes D_{n-p}$ in $(C\otimes D)_n$. Then the map extends by linearity. $\endgroup$
    – user113102
    Oct 4, 2019 at 14:11

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