Homomorphism of chain complexes induces homotopy/isomorphism of chain complexes I was given the following exercise, but the person who gave me the exercise wasn't sure about some of the details (such as signs).
Let $(C, \partial)$ be a chain complex, and $h: (C, \partial) \rightarrow (C, \partial)$ satisfy $h^2 = 0$. Then $(C, \partial) \cong (C, \partial \pm h\partial \pm \partial h)$. 
I am not sure if $\cong$ is supposed to be an isomorphism or a chain homotopy. The symbol suggest isomorphism, but I am not sure if we are working in the category of chain complexes up to homotopy.
Using $h^2 = 0$ gives us that $(C, \partial \pm h\partial \pm \partial h)$ is indeed a chain complex. If the signs on $h\partial$ and $\partial h$ differ, then the isomorphism is trivial as $\partial h = h \partial$. Thus I believe that the signs are both $+$. If somebody knows precisely the statement of the problem, that would help me do the exercise. 
 A: I don't know the precise statement but here are some thoughts on how to attack a statement that you don't know precisely : 
you can try to fiddle around with things and see what you get; for instance I suspect the statement is an isomorphism, because the given data doesn't allow you to define anything $C_n\to C_{n+1}$. So if I were in your situation I'd try to see if I can prove an isomorphism, and if not, what's the obstruction ? Answering this will  probably give you the homotopy if it's not an isomorphism.
So try to see what an isomorphism $f$ would have to satisfy : first of all, it has to be a morphism of chain complexes : what equation does that give you ? Can you see easy nonzero solutions to that equation ? 
An isomorphism should "look like" the identity, so you can assume $f$ is of the form $id_C+$some perturbation, and see what that perturbation ought to look like, etc. To sum up : in this situation, try to fiddle around with the data, and often you'll find the correct statement and the proof at the same time.
