Signs in the tensor product and internal hom of chain complexes

Solution 1:

One algebraic way to motivate this is to observe that the signs in the differential for the Hom are precisely what is needed for 0-cycles in the $\hom(A,B)$ complex to be the set of morphisms of complexes $A\to B$ (and also, that the 0th homology group $H_0(\hom(A,B))$ is the set of homotopy classes of morphisms $A\to B$). This is quite great.

Once you decide you want this, all the other signs you mention follow because you need various things to hold. For example, you want the adjuntion between $\hom$ and $\otimes$ to hold for the internal versions, so this forces you to add signs to the $\otimes$, and so on.

A topological, indirect explanation for the appearence of most signs is that in the long exact sequence of bases maps corresponding to a map $f:X\to Y$, which looks like $$X\to Y\to C(f)\to S(X) \to S(Y) \to S(C(f)) \to \cdots$$ there is a "sign" which you cannot get rid of. This sign reproduces itself in every algebraic version.