Chain Homotopy classes as n-homology of a double complex Hey all,
I try to understand an argument in Lücks "A basic introduction to surgery theory" on page 51 which goes as follows:
Let $\mathbb{Z} \pi$ be the group ring where $\pi$ denotes the fundamental group of a finite connected CW-Komplex. Furthermore we regard the zellular $\mathbb{Z}\pi$-chain complex $C_*(\widetilde X) $ of the universal covering of $X$.
Now it is claimed, that taking the n-th homology of the double complex $hom(C^{-*}(\widetilde X),C_*(\widetilde X))$ is the set of $\mathbb{Z}\pi$-chain homotopy classes of $\mathbb{Z}\pi$-chain maps from $C^{n-*}(\widetilde X)$ to $C_*(\widetilde X) $.
Here $C^{n-*}(\widetilde X)$ denotes the dual chain complex given by
$C^{n-*}(\widetilde X)_p:=hom(C_{n-p},\mathbb{Z}\pi) $ with the boundary map $ \circ \partial^C $.
If I regard the n-th boundary map of the double complex $hom(C^{-*}(\widetilde X),C_*(\widetilde X))$ I considered as definition of the n-th boundary map
$\partial^n f_p :=f_p \circ\partial^{C^{-*}}+(-1)^{n-1}\partial^{C_*} \circ f_{p+1}$ ? This seems to be almost the definition of a chain homotopy but the prefactor $(-1)^{n-1} $ causes troubles.
I would be grateful for any references to literature.
 A: Part of the problem is that you're probably only used to seeing chain homotopies of degree zero chain maps. In this case $n=1$ (because the homotopy itself raises degree by 1 and so is an element of $Hom^1$) and the formula reduces to the standard formula for a chain homotopy that you'd find, for example, in Hatcher or Munkres or any other introduction to algebraic topology. In this more general setting, elements of $Hom^n(C^\ast, D^\ast)$ (I'm generalizing the setting to pure algebra - though note that algebraically $C^\ast=C_{-\ast}$ by convention) raise degree by $n$ and certain signs come in when you work with homomorphisms of chain complexes that shift degrees like this. 
Probably the best explanation for why you want the signs in the general case is the explanation I first found here in Section 12 of Lawson's paper about signs (though no doubt many people have known this for a long time). The point is that given the sign conventions for tensor products (which are also reasonable from the theory of symmetric monoidal categories) this is the only choice of sign convention that will make the evaluation map $Hom(C^\ast, D^\ast)\otimes C^\ast\to D^\ast$ a chain map, which is a pretty reasonable demand. Try to show yourself that this evaluation is a chain map and you'll see where the mysterious sign comes into it. 
A: It took me quite a while to figure out the correct definitions but now there is only one "mysterios sign" left. To be more precised: 
Given two $R$-Chain complexes $C_{*}$, $D_{*}$ we regard the complex $hom_R(C_{*},D_{*})$ whose p-th module is given by $hom_R(C_{*},D_{*})=\bigoplus_k hom_R(C_{k-p},D_k)$ and boundary map $d^{p}(f):=d_Df-(-1)^pfd_C$. If you now regard the shifted complex $\Sigma^{n} C_{*}$ with $(\Sigma^{n} C_{*})_{k}:=C_{k-n}$ you "have to" define it's boundary map by $d_{\Sigma C}:=(-1)^{n} d_{C} $.Using this definitions it is easy to show that 
$H_{n}(hom_{R}(C_{*},D_{*}))=[\Sigma^{n} C_{*},D_{*}] $, where $[.,.]$ denotes chain homotopy classes as above.
So maybe the question remaining is, why you need the $(-1)^n$ in the boundary map for the shifted complex.
