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Let $R$ be an associative ring with unit. Let $E$ and $F$ be two chain complexes of $R$-modules and $\phi: E\overset{\sim}{\to} F$ be a quasi-isomorphism between them, i.e. $\phi$ induces ismorphisms on cohomology modules at each degree.

We call $\psi: F\to E$ a left homotopy inverse of $\phi$ if there exist a degree $-1$ map $\eta: E\to E$ such that $\psi\circ \phi-\text{id}_E=d \eta+\eta d$. Similarly we call $\psi: F\to E$ a right homotopy inverse of $\phi$ if there exist a degree $-1$ map $\tau: F\to F$ such that $\phi\circ \psi-\text{id}_F=d \tau+\tau d$.

My question is: if we know $\phi: E\overset{\sim}{\to} F$ is a quasi-isomorphism and $\psi: F\to E$ is a left homotopy inverse of $\phi$, does it automatically mean that $\psi$ is also a right homotopy inverse of $\phi$? or vice versa?

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  • $\begingroup$ Isn't it $d\eta + \eta d$ and $d\tau + \tau d$ ? $\endgroup$ – Max Jan 11 at 20:49
  • $\begingroup$ @Max Yes I have modified it. $\endgroup$ – Zhaoting Wei Jan 12 at 1:45
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The answer is no with the usual notion of homotopy (which is not the one you gave in your question so I don't know whether that'll answer your question - it will if it was a typo)

Indeed consider a complex that has zero homotopy but is not contractible, say $E=(0\to I\to R\to R/I\to 0)$ for $I$ a proper ideal of $R$ ($R$ commutative)

Then the unique map $f:E\to F$, where $F$ is the $0$ complex, is a quasi-isomorphism, and if $g$ denotes the unique map $F\to E$, we have that $f\circ g= id_F$ so $f$ is a left homotopy inverse of $g$ but $g\circ f= 0$ is not homotopic to $id_E$ (by choice of $E$) so that $f$ is not a right homotopy inverse of $g$.

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