Set $f:Y \to X$ be a morphism between schemes and $s \in \Gamma(X, \mathcal{O}_X)$ be a global section.

The map $f$ induces a functor $f^*$ (so called pullback functor) that pulls back $\mathcal{O}_X$-modules over $X$ to $\mathcal{O}_Y$ modules.

Futhermore for the global section $s$ I often encounter the notation $f^*s$ ("pullback of a global section")

Using the well know adjunction correspondence between pullback and pushforward we get for arbitrary sheaf $\mathscr F$ a natural morphism of sheaves $\mathscr F\to f_* f^*\mathscr F$. Obviously this induces in functorial sense a map between global sections

$$H^0(X,\mathscr F)\to H^0(X,f_* f^*\mathscr F)=H^0(Y,f^*\mathscr F)$$

which exactly maps $s$ to $f^*s$.

Formally that's ok. My problem is how it concretely looks like.

The most easiest case that $X= Spec(R), Y= Spec(A)$ and $f$ is induced by the ring map (=map between global sections) $\varphi_f: R \to A$.

Therefore every $r \in R$ is a global section.

How concretely in this case the pullback $f^*r \in A$ looks like and how to derive/conclude it?

  • 3
    $\begingroup$ $f^*r = \phi_f(r)$. This is really how the pullback is defined if you go back to the definition. $\endgroup$ – Nicolas Hemelsoet Dec 26 '18 at 22:26
  • $\begingroup$ @NicolasHemelsoet: Ok, let's see. In affine case the adjunction formula reduces to following: Let $M$ be a $R$-module (the pendant to the $\mathcal{O}_Y$-module- later M=R - then we have the bijection $$Hom_A(M \otimes A, M \otimes A)= Hom_R(M, f_*(M \otimes A))$$ (take into account $f^*M= M \otimes A$) By construction $b: M \to f_*(M \otimes A) $ comes from $id_{M \otimes A}$ on the left side. Now assume $M= R$. $\endgroup$ – KarlPeter Dec 26 '18 at 22:53
  • $\begingroup$ So $b: R \to f_*(A)$. Why $f_*(A) = A$? I don't see how the pushforward functor $f_*$ acts on level of rings/ local sections. And then why $b= \varphi_f$? How the counit concretely maps the identity to the right set? $\endgroup$ – KarlPeter Dec 26 '18 at 22:54
  • $\begingroup$ If $f:X=Spec(A)\to Spec (B)=Y$ is a morphism then $f_*M$ for $M$ being a $A$-module is simply $M$ seen as a $B$-module via the map $\phi: B \to A$ corresponding to $ f^{\#} : \Gamma(Y, O_Y) \to \Gamma(Y, f_* O_X)=\Gamma(X, O_X) $ (in other words the map of ring $B \to A$ that defines $f$). One possible way to see this (but not the only one, but given what you wrote it is possibly the one that will satisfy you the most?) is to note that adjunction property $Hom_A(N\otimes_B A, M)=Hom_B(N,M_B)$ where $M_B$ is simply $M$ viewed as a $B$-module. Yoneda lemma ensures that $M_B=f_*M$ $\endgroup$ – Ahr Dec 27 '18 at 10:11
  • $\begingroup$ If you want to consider the problem in more geometric (kinda) terms, you might note that $f^{-1}(D(b))=D(\phi(b))$ and that two quasi coherent sheaves coincide iff they coincide on principal open subsets. All you have to do now is compare the quasi coherent sheaf defined by $M_B$ and the quasi coherent sheaf $f_*M$ on those open subsets, but by what is said $\tilde{M_B}(D(b))$ is $M_b$ where the $B$-module structure is given by $\phi$ and $f_*M(D(b))=M_{\phi(b)}$. This is the same thing. $\endgroup$ – Ahr Dec 27 '18 at 10:22

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