# 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.

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.