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 $(\phi_n)$?


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.

  • $\begingroup$ But what I'm really asking is the relation between chain homotopy equivalence in the category of chain complexes and the corresponding isomorphisms in the category of abelian groups. Is this relation also false? $\endgroup$ – user42383 Nov 16 '14 at 4:27
  • $\begingroup$ @user42383: yes, that's also false. If a chain map has an inverse levelwise then the levelwise inverses are necessarily also a chain map giving an inverse in the strict sense; those conditions are equivalent and are not equivalent to being a chain homotopy equivalence. Have you looked at any examples? $\endgroup$ – Qiaochu Yuan Nov 16 '14 at 4:30
  • $\begingroup$ Maybe I ask what the definition of an inverse of a chain map? $\endgroup$ – user42383 Nov 16 '14 at 15:54
  • $\begingroup$ @user42383: in any category, an inverse of a morphism $f$ is a morphism $g$ such that $f \circ g$ and $g \circ f$ are both identity morphisms. A chain map is a morphism in the category of chain complexes. $\endgroup$ – Qiaochu Yuan Nov 17 '14 at 4:27

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