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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?

user267839
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    $f^*r = \phi_f(r)$. This is really how the pullback is defined if you go back to the definition. – Nicolas Hemelsoet Dec 26 '18 at 22:26
  • @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 $M= R$.

     – user267839 Dec 26 '18 at 22:53
  • 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? – user267839 Dec 26 '18 at 22:54
  • 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 $M_B=f_*M$ – Ahr Dec 27 '18 at 10:11
  • 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. – Ahr Dec 27 '18 at 10:22
  • @A.Rod: So this shows that locally (or in affine case) $f_M = M \vert B$ is exactly the restriction. What I can't deduce from this is why this already imply that the canonical map $b: M \to f f^* M $ that comes via adjunction from $id_{m \otimes A}$ is - in case $M= B$ - imduce exactly $\phi$ (or $f^{#}$ using scheme formalism).

    Indeed $b$ induce as a sheaf morphism also a map $b^{#} : \Gamma(Y, O_Y) \to \Gamma(Y, f_* O_X)=\Gamma(X, O_X)$ between global sections but why does it coinside with $f^{#}=\phi$ ?

     – user267839 Dec 28 '18 at 00:11
  • Naively one can think that $b^{#}$ and $f^{#}= \phi$ just two different maps $B \to A$. – user267839 Dec 28 '18 at 00:11
  • @A.Rod: One further remark: What is $f^{#}$ by definition in case of affine schemes as above? Do you mean that $f^{#} = \phi$ by definition (that's my actual point of view) or do you take as rough definition for $f^{#}$ the induced global section map of our trusty friend $\mathcal{O}Y \to f* f^*\mathcal{O}_Y$ from above.

    I know that indeed the both SHOULD coinside but why is exactly the core problem of mine/this thread since the later is exactly the map $s \mapsto f^*s$

     – user267839 Dec 28 '18 at 00:12
  • I'll answer your last question 1st, the map $f^#$ is the map induced on global section and also the morphism of rings $\phi: B \to A$ that defines $f$. The fact that it's the same morphism comes from the antiequivalence between the category of affine schemes and the category of rings. Indeed you know that if $f: X \to Y$ is a morphism of affine schemes, you get back the rings by noting that $A=O_X(X)$, $B=O_Y(Y)$ and the morphism $f$ coincide with $Spec(f^#)$ with $f^#: B \to A$ on the global sections, this is indeed a theorem that relies on the mac that morphisms of schemes are local. – Ahr Dec 28 '18 at 08:45
  • This is done at the beginning of every textbook on schemes. To understand why the map you're looking at is what it is, recall that it is this $f^#$ that gives the $B$-algebra (and B-module) structure on $A$. The map $Hom_A(M\otimes A, N)\simeq Hom_B(M, N_B)$ is simply given by "extension by $A$-linearity" i.e $h(a\otimes m)=ah(m)$ for any $h: M \to N_B$. – Ahr Dec 28 '18 at 09:04
  • Since the map giving the $B$-module structure on $A$ is $\phi=f^#$, this map in $Hom_B(B, A)$ is mapped to $a\otimes b \mapsto a\phi(b)$ this is extactly the map providing the isomorphism $B\otimes_B A \simeq A$, thus it is mapped to the identity in $Hom_A(A, A)$ via the map $Hom_A(B\otimes_B A, A)\simeq Hom_A(A,A)$ given by composition with the above mentionned isomorphism. – Ahr Dec 28 '18 at 09:04
  • @A.Rod: Yes, seems become clearer but till now I see one unclear point. Let me summarize: For affine case you describe the adjunction bijection $Hom_A(M\otimes A, N)\simeq Hom_B(M, N_B)$ is given concretely by maping a $h: M \to N_B$ to map on the left side with property $h(a\otimes m)=ah(m)$. At this point: Do you maybe have a reference/ sketch of an argument why this map is exactly the adjunction bijection which arises as affine special case of the bijection $$Hom_{\mathcal{O}X}(f^*\mathcal{G},\mathcal{F})= Hom{\mathcal{O}Y}(\mathcal{G}, f*\mathcal{F})$$? – user267839 Dec 29 '18 at 02:14
  • @A.Rod: Sure this is bijective and is constructed quite canonically but up to now that offers to be a huge problem to find literature with concrete description of the adjucntion morphism. From now let assume that this is exactly the map. Then the inverse map should be (I guess) the restriction of a map with domain $M \otimes A$ to elements of the shape $m \otimes 1$ and forget the $A$-action. Take in our case $M=B, N =A$. – user267839 Dec 29 '18 at 02:15
  • And then - if I understood your arguments correctly - you show that $\phi$ and the other map (the protagonist of this thread) corresponding by bijectivity exactly to $id_A$, so they must coinside. Great, so up to the point I mentioned above that's clear. – user267839 Dec 29 '18 at 02:15
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    Does this answer your question? What is a simple definition of the pull-back of a section? – Manos Jan 24 '20 at 16:03

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