**Hatcher Exercise 2.1.12**

**Theorem**: Chain homotopy of chain maps is an equivalence relation.

*Proof*:
Suppose we have chain maps .
For and to be *chain-homotopic*, which we will denote for the duration of this proof by , means that there exists a map such that (for all ).
We will abuse notation and and just write the left side as , leaving the particular instance (subscripts) of and to be inferred from the context.

Clearly, is reflexive. Simply take to be the zero map, and we have . So, .

It’s also easy to see that is symmetric. Suppose so that for some . Then, we have . So via the map .

Finally, suppose and . Then, we have and for maps and . Then,

This last line perhaps merits a bit of explanation – since these are all maps between abelian groups, addition of maps (defined in the obvious way) is clearly commutative, and the fact that can factor out of sums follows from the fact that is a homomorphism. So, via the map .

So, chain-homotopy is an equivalence relation.