If $f \circ g$ and $f$ are quasi-isomorphisms, is $g$ a quasi-isomorphism ?
Best Answer
Yes, because (co-)homolgy in degree $n\in\mathbb N$ is a functor and compatible with composition:
Let $g:C\Rightarrow D$ and $f:D\Rightarrow E$ be chain-maps such that $f\circ g$ and $f$ are quasiisomorphisms. By definition for any $n\in \mathbb N$ we have that $H_n(f\circ g)=H_n(f)\circ H_n(g)$ and $H_n(f)$ are isomorphisms. Then $H_n(g)=H_n(f)^{-1}\circ H_n(f\circ g)$ is an isomorphism as the composition of two isomorphisms. So for all $n\in \mathbb N$ we have that $H_n(g)$ is an isomorphism, that is $g$ as quasiisomorphism.
$\Leftarrow$: if $F$ is exact, then it commutes with taking homology; in particular, it preserves acyclic complexes. Now use the fact that a map is a quasi-isomorphism iff its mapping cone is acyclic and that any functor on chain complexes induced from a functor between abelian categories preserves mapping cones.
$\Rightarrow$: we'll prove the contrapositive. Suppose $0 \to a \to b \to c \to 0$ is a short exact sequence not preserved by $F$, so $F(a) \to F(b) \to F(c)$ is not exact. Now, a three-term complex is exact iff it's acyclic iff it's quasi-isomorphic to $0$, so $F$ does not preserve quasi-isomorphisms.
No, they needn't be homotopy equivalent. Let $$X = \dots \rightarrow \Bbb Z/4\Bbb Z \xrightarrow{2\times} \Bbb Z/4\Bbb Z\xrightarrow{2\times} \Bbb Z/4\Bbb Z \rightarrow \dots$$ and $Y$ be the zero complex. Then $X$ and $Y$ have trivial homology and the zero maps between them are quasi-isomorphisms; but the identity map on $X$ is not null-homotopic, so that $X \not\sim Y$.
Best Answer
Yes, because (co-)homolgy in degree $n\in\mathbb N$ is a functor and compatible with composition:
Let $g:C\Rightarrow D$ and $f:D\Rightarrow E$ be chain-maps such that $f\circ g$ and $f$ are quasiisomorphisms. By definition for any $n\in \mathbb N$ we have that $H_n(f\circ g)=H_n(f)\circ H_n(g)$ and $H_n(f)$ are isomorphisms. Then $H_n(g)=H_n(f)^{-1}\circ H_n(f\circ g)$ is an isomorphism as the composition of two isomorphisms. So for all $n\in \mathbb N$ we have that $H_n(g)$ is an isomorphism, that is $g$ as quasiisomorphism.