On chain homotopy equivalence

Question

I just learnt the notion of chain map and have the following question. Let $C=(C_n,\partial_n^C)$ and $D=(D_n,\partial_n^D)$ be chain complexes of abelian groups with boundary maps $\partial_n^C$ and $\partial_n^D$. Then I guess it should be true that $\phi:C\to D$ is a chain homotopy equivalence precisely when $\phi_n:C_n\to D_n$ is an isomorphism for all $n$. $$\begin{array} $C_{n+1} & \stackrel{\phi_{n+1}}{\longrightarrow} & D_{n+1} \\ \downarrow{\partial_{n+1}^C} & & \downarrow{\partial_{n+1}^D} \\ C_{n} & \stackrel{\phi_{n}}{\longrightarrow} & D_{n} \end{array} $$ One direction seems easy: if $\phi=(\phi_n)$ is a chain homotopy equivalence, then there is a homotopy inverse $\psi=(\psi_n)$, so $\psi_n$ is an inverse of $\phi_n$ in the category of abelian groups. But I don't know how to show the converse, i.e., how to show $(\phi_n^{-1})$ is the homotopy inverse of $(\phi_n)$?

Answer 1

No, that's not true. A chain homotopy equivalence is not chain map with an inverse; it is something weaker (namely it is a chain map with an "inverse up to chain homotopy," exactly the way it sounds). In particular you're confused about which direction is easy: the easy direction is that if a chain map has an inverse then it is a chain homotopy equivalence. The other direction is false.