Let $(B, d_B),(C, d_C),(D, d_D)$ and $(E, d_E)$ be cochain complexes and let $f, g: C \rightarrow D$ be chain homotopic maps. I would like to prove the following claim:
If $u: B \rightarrow C$ and $v:D \rightarrow E$ are cochain maps, then $vfu$ and $vgu$ are chain homotopic.
Since $f \sim g$, by definition there is a homotopy $h: C \rightarrow D$ such that $f-g = hd_C + d_Dh$. Furthermore, as $u$ and $v$ are cochain maps, they commute with the differentials, i.e. $d_Cu = ud_B$ and $d_Ev = vd_D$. I need to show that there is a homotopy $h:B \rightarrow E$ such that $vfu - vgu = hd_B + d_Eh$. Can someone explain to me how to proceed ?
Thanks for your help.
Take as a homotopy $vhu \colon B \to E$. Now check that it works:
Let $b\in B$. Then $[vfu-vgu](b)=v(\alpha)$ where $$ \alpha=[f-g](u(b))=[hd_{C}+d_{D}h](u(b))=hd_{C}u(b)+d_{D}hu(b)=hud_{B}(b)+d_{D}hu(b)$$ and then applying $v$ and using as in the last equality the property of chain maps we finally get $$ [vfu-vgu](b)=vhud_{B}(b)+d_{E}vhu(b) $$