Given a chain complex $(C_n,\partial_n)$ of free modules over $R$ PID, we can apply the $\operatorname{Hom}(-,R)$ functor to induce a cochain complex $(C^n,\delta^n$) where $C^n :=\operatorname{Hom}(C_n,R)$ and $\delta^{n-1}\colon\operatorname{Hom}(C_{n-1},R) \longrightarrow \operatorname{Hom}(C_n,R)$ defined throught the old differential, i.e $\delta^{n-1}(\varphi) = (-1)^n \varphi \circ \partial_n$.
Given the classical definition of chain homotopy how the following statements could be true :
Lemma : If $f$ is chain equivalence then $^tf$ is chain equivalence
The argument reduces in showing that if $f \simeq g$ through certain $s$ then $^tf \simeq \hspace{0.1cm}^tg$ through $^ts$, but when I try to compute directly the terms I get have different parity, which doesn't allow me to get the "classic" formula : $^ts \partial + \partial ^ts = ^tf - ^tg $.
Any idea on how should I proceed ? Any help would be appreciated.
A chain homotopy has degree $1$ so the dual homotopy should satisfy $s(\phi) = (-1)^n \phi\circ s$ where $\phi: C_n \to R$. In this case you can see that $$ (s\partial + \partial s)(\phi) = (-1)^{n+1} s(\phi \circ \partial) + (-1)^n \partial(\phi\circ s) = (-1)^{n+1+n+1}\phi\circ\partial\circ s + (-1)^{n+n} \phi\circ s\circ \partial =\phi(\partial\circ s + s\circ\partial) = \phi(f-g) = \phi(f)-\phi(g) $$ Trust in Koszul.