Suppose I have an arrow $\varphi :f^{-1}G\rightarrow F$ which is a map of $f^{-1}\mathcal O_Y$-modules, i.e such that the square below commutes $$\require{AMScd} \begin{CD} f^{-1}\mathcal O_Y\otimes f^{-1}G @>>> f^{-1}G\\ @V{1\otimes \varphi}VV @VV{\varphi}V\\ f^{-1}\mathcal O_Y\otimes F @>>> F \end{CD}$$
How can I show that its transpose $\psi:G\rightarrow f_\ast F$ is a map of $\mathcal O_Y$-modules, i.e that the square below commutes? $$\require{AMScd} \begin{CD} \mathcal O_Y\otimes G @>>> G\\ @V{1\otimes \psi}VV @VV{\psi}V\\ \mathcal O_Y\otimes f_\ast F @>>> f_\ast F \end{CD}$$
All I know is that the map $\mathcal O_Y\otimes f_\ast F\rightarrow f_\ast F$ which makes $f_\ast F$ into an $\mathcal O_Y$-module is given by the composite $\mathcal O_Y\otimes f_\ast F\overset{f^\sharp \otimes 1}{\longrightarrow}f_\ast \mathcal O _X \otimes f_\ast F\longrightarrow f_\ast F$. Here, $f^\sharp$ is part of the morphism of ringed spaces $X,Y$.