Recall that a chain complex is a (finite) diagram of the form $$V = \{ \dots \to V_3 \overset{d_3}\to V_2 \overset{d_2}\to V_1 \overset{d_1}\to V_0 \to 0 \}$$ where the $V_n$ are (finite-dimensional) vector spaces and for each $n$, $d_n \circ d_{n+1} = 0$. If $V$ and $W$ are chain complexes, a chain map $f: V \to W$ is a map $f_n : V_n \to W_n$ for each $n$ such that all the obvious squares commute — "$[d,f]=0$" — and the pair (chain complexes, chain maps) defines a category. In fact, it is a 2-category: the 2-morphisms between $f,g : V \rightrightarrows W$ are the chain homotopies, i.e. a system of maps $h_n: V_n \to W_{n+1}$ such that "$[d,h] = f-g$". The category of chain complexes has a biproduct (both a product and a coproduct) $\oplus$ given by the pointwise direct sum.
I thought I knew what the tensor product of chain complexes was. Namely, if $V$ and $W$ are chains, then the usual thing is to define $$(V\otimes W)_n = \bigoplus_{k=0}^n V_k \otimes W_{n-k}$$ and the chain maps are the sums of the obvious tensor products of differentials, decorated with signs.
But now I'm not sure why this is the tensor product picked. Namely, if I have a linear category, I think that a tensor product $V \otimes W$ should satisfy the following universal property: for any $X$, $\hom(V \otimes W,X)$ should be naturally isomorphic to the space of bilinear maps $V \times W \to X$. Now, I've never really known how to write down the word "bilinear" in a general category, without refering to individual points. But I think I do know what the "set" $V \times W$ is when $V$ and $W$ are chains — it's the set underlying $V \oplus W$ — and then I think I do know what bilinear maps should be.