Is a left homotopy inverse of a quasi-isomorphism automatically a right homotopy inverse?

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?

• Isn't it $d\eta + \eta d$ and $d\tau + \tau d$ ? – Max Jan 11 at 20:49
• @Max Yes I have modified it. – Zhaoting Wei Jan 12 at 1:45

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 $$g$$.