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Let $(f, f^\#): X \to Y$ a morphism between schemes, therefore $f: X \to Y$ between underlying topol spaces and $f^\#:\mathcal{O}_Y \to f_*\mathcal{O}_X$ sheaf morphism. Let $s \in \Gamma(Y,\mathcal{O}_Y)$ be a global section. Obviously $s$ corresponds uniquely to a sheaf morphism, denoted by $\hat{s} : \mathcal{O}_Y \to \mathcal{O}_Y$.

My question is why the following diagram commutes:

$$ \require{AMScd} \begin{CD} f^* \mathcal{O}_Y @>{f^* \hat{s}} >> f^* \mathcal{O}_Y \\ @VVV @VVV \\ \mathcal{O}_X @>{\widehat{f^\#(s)}}>> \mathcal{O}_X \end{CD} $$

Remark: Vertical arrows are canonical isomorphisms.

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Are you really too lazy to check that the diagram $$ \require{AMScd} \begin{CD} A \otimes_B B @>{a \otimes b \mapsto sa \otimes b} >> A \otimes_B B \\ @V{a \otimes b \mapsto \varphi(a)b}VV @VV{a \otimes b \mapsto \varphi(a)b}V \\ B @>{\cdot \varphi(s)}>> B \end{CD} $$ commutes for a ring map $\varphi: A \to B$?

Either path sends $a \otimes b$ to $\varphi(s)\varphi(a)b$.

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  • $\begingroup$ Haha, sorry, in light of commutative algebra the question seems indeed ridiculous. The problem was to translate the morphism $f^* \hat{s}$ to the ring side. $\endgroup$
    – KarlPeter
    Mar 20 '18 at 12:22

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