I am currently studying some homological algebra, I have a couple of questions concerning the notion of quasi-isomorphism and homotopical equivalence.
For two complexes on an abelian category $X^\bullet$ and $Y^\bullet$ to be homotopically equivalent means that there are morphisms $u:X^\bullet\rightarrow Y^\bullet$ and $v:Y^\bullet\rightarrow X^\bullet$ such that $v\circ u \sim 1_{X^\bullet}$ and $u\circ v \sim 1_{Y^\bullet}$ (where by "$\sim$" I mean the homotopical equivalence for morphisms).
In this case one can prove that $u$ and $v$ are quasi-isomorphism. I read that the converse is false in the sense that if there exists a quasi-isomorphism $u:X^\bullet \rightarrow Y^\bullet$ then $X^\bullet$ may not be homotopical equivalent to $Y^\bullet$. A counterexample is this: https://math.stackexchange.com/a/93284/142179. But what happens if we have two quasi-isomorphisms $u:X^\bullet \rightarrow Y^\bullet$ and $v:Y^\bullet \rightarrow X^\bullet$? Is it true then that $X^\bullet \sim Y^\bullet$?
I have also a softer question: what is a topological analogous of a quasi-isomorphism? I mean, is there a "property" that two continuous functions $u:X\rightarrow Y$ and $v:Y\rightarrow X$ have to satisfy if $v\circ u \sim \text{id}_X$ and $u\circ v \sim \text{id}_Y$?
Best Answer
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 $X \not\sim Y$.