# Understanding the definition of a tensor product of chain complexes of abelian groups

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

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.

• The key is to not write a bunch of diagrams and induced maps but rather to just explicitly write out what the map is. – Connor Malin Oct 3 at 21:38

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 $$g_{n,p}$$.
• 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. – Pel de Pinda Oct 3 at 23:40
• 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. – user113102 Oct 4 at 14:11