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Let $\varphi \colon X \rightarrow Y$ be a morphism of varietys, $\mathcal{F}$ a locally free sheaf of $\mathcal{O}_{Y}$-modules and $L$ the associated vector bundle. Then we can construct the inverse image sheaf $\varphi^* \mathcal{F} = \varphi^{-1} \mathcal{F} \otimes_{\varphi^{-1} \mathcal{O}_{Y}} \mathcal{O}_{X}$. Furthermore there is the so called pullback bundle $\varphi^*L = L \times_Y X$.

According to,"the pullback of bundles corresponds to the inverse image of sheaves" but how can I see this?

Thanks in advance!

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The way I think about this (and it should be isomorphic to the way any reputable source does) is the following:

$L$ is locally equal to $\operatorname{Spec} S(\mathcal{F})$, the spectrum of the symmetric tensor algebra over $\mathcal{F}$. This is pretty easy to understand on an affine patch—you are taking formal sums of tensors of module elements $a+b\otimes c + d\otimes e\otimes f \otimes g + u\otimes w \ldots$, where you quotient out by $x\otimes y - y\otimes x$ so that it's commutative.

Since $\mathcal{F}$ is locally free, it's not too hard to see that $S(\mathcal{F})$ is locally isomorphic—over an open affine patch $\operatorname{Spec} A=U\subset Y$ small enough to trivialize $\mathcal{F}$—to $A[x_1,\ldots,x_n]$, where $n$ is the rank of $\mathcal{F}$.

Let me make that explicit. On $U$, $\mathcal{F}$ looks like the module $M=\bigoplus_{i=1}^n Ax_i$. Then $S(M)$ is exactly the free algebra over $A$ generated by $x_1, \ldots , x_n$. So $S(M) = A[x_1,\ldots,x_n]$, and we can see that it's reasonable to call $\operatorname{Spec} S(\mathcal{F})$ a "vector bundle", since locally it looks like $\mathbb{A}^n_U$.

Okay, now let's suppose that $X\to Y$ is locally given by a morphism of rings $A\to B$, where $\operatorname{Spec} B = V \subset X$. The pullback of $M$ is clearly the base change $M\otimes_A B = \bigoplus_{i=1}^n B x_i$.

But how about pulling back the bundle? Well, it's just $\mathbb{A}^n_U \times_U V = \mathbb{A}^n_V$. Or, on the level of rings, we have $S(M) \otimes_A B = S(M\otimes_A B)$. And that works at the sheaf level, too: $S(\mathcal{F}) \otimes_{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X = S(\mathcal{F}\otimes_{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X)$.

So this all adds up to a pretty strong dictionary (actually, an equivalence of categories) between locally free sheaves and so-called "geometric vector bundles". Actually, it's a little more general than that—if you look closely at how morphisms of sheaves lift up to the corresponding bundles, you find that as long as $\mathcal{F}$ is reflexive—that is, the natural map $\mathcal{F}\to\mathcal{F}^{\vee\vee}$ is an isomorphism—it can pretty much be identified with its bundle—the sheafy $\operatorname{Spec} S(\mathcal{F})$. You can think of these as singular bundles.

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Your answer made this a lot clearer for me. But there remains the following question: The pullback of the bundle locally looks like $\mathbb{A}^n_U \times _U V= Spec(S(M)) \times _{Spec(A)} Spec(B)$. To answer my question I think I should show that this is locally the same as $Spec(S(\varphi^{-1}{F} \otimes _{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X))$ but I don't know how to do this. – claudi Oct 18 '13 at 11:36
@claudi It's really about the simpler-looking $S(M)\otimes_A B = S(M\otimes_A B)$. You could do this with universal properties, or you could look at $S(M)$, as a module, as the direct sum $\bigoplus_m M^{\otimes m}$. – Slade Oct 18 '13 at 17:37

There are lots of nice answers to this question already, and this may not add much that isn't already present, but just to emphasize, there's no reason to analyze things affine locally, as this is a consequence of largely formal properties of the construction of the equivalence (anti-equivalence if you choose to omit the dual) between geometric vector bundles and finite locally free $\mathscr{O}_X$-modules via the relative spectrum (on any scheme $X$ whatsoever). Namely once one constructs the relative spectrum of a quasi-coherent algebra (which at some point does require one to work with affine opens), the compatibility with pullback follows from whatever characterizing property one chooses to use for the relative spectrum (my personal preference is for using the functor it represents).

If $\mathscr{F}$ is any finite locally free $\mathscr{O}_X$-module, the corresponding geometric vector bundle is $\mathbf{A}(\mathscr{F})=\underline{\mathrm{Spec}}_X(\mathrm{Sym}_{\mathscr{O}_X}(\mathscr{F}^\vee))$ (I'm not totally sure if this is the standard convention or if omitting the dual is the standard convention). In general, if $\mathscr{A}$ is a quasi-coherent $\mathscr{O}_X$-algebra (such as $\mathrm{Sym}_{\mathscr{O}_X}(\mathscr{F}^\vee))$, and $g:X^\prime\to X$ is a morphism, then there is a canonical isomorphism $\underline{\mathrm{Spec}}_{X^\prime}(g^*(\mathscr{A}))\simeq X\times_X\underline{\mathrm{Spec}}_X(\mathscr{A})$ (one can verify this very easily using the description of the functor of points of the relative spectrum which is sometimes taken as its defining property). In the vector bundle case, one also uses that pullback commutes with formation of the symmetric algebra (and duality, if you include that dual) for finite locally free modules (formation of the symmetric algebra commutes with pullback for general quasi-coherent modules, not just finite locally free ones). For details about these things, I recommend and, as well as Chapter 11 of the wonderful book by Görtz-Wedhorn.

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$Sh(X)\leftrightarrows Et(X)$ is the deep reason; for example in "Sheaves in Geometry and Logic" the inverse image sheaf is defined as the sheaf associated to the pullback of the bundle which corresponds to the sheaf you're considering.

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For simplicty, I start from the case of the trivial vector bundle over $\operatorname{Spec}\mathbb{K}$.

Let $\mathbb{V}$ be a vector space over a field $\mathbb{K}$ of finite dimension; defined \begin{equation*} \mathbb{K}[\mathbb{V}^{\lor}]=\bigoplus_{n\in\mathbb{N}}S^n(\mathbb{V}^{\lor})\equiv\operatorname{Sym}(\mathbb{V}) \end{equation*} where $\mathbb{V}^{\lor}$ is the dual space and $S^n(\cdot)$ is the $n$-th symmetric power of $\cdot$; we get \begin{equation*} \mathscr{V}=\operatorname{Spec}\mathbb{K}[\mathbb{V}^{\lor}], \end{equation*} because \begin{gather} \forall A\in Ob(\mathbf{Alg}_{\mathbb{K}}),\,\mathscr{V}(\operatorname{Spec} A)=\hom_{\mathbf{Sch}}(\operatorname{Spec}A,\mathscr{V})\cong\hom_{\mathbf{Alg}_{\mathbb{K}}}(\mathbb{K}[\mathbb{V} ^{\lor}],A)\cong\\ \cong\hom_{\mathbf{Mod}_{\mathbb{K}}}(\mathbb{V}^{\lor},A)\cong\mathbb{V}\otimes_{\mathbb{K}}A; \end{gather} where the categories are obvious.

In particular, if $\dim\mathbb{V}=m$, one can prove that: \begin{equation} \mathscr{V}\cong\mathbb{A}^m_{\mathbb{K}}, \end{equation} that is $\mathscr{V}$ is the trivial (geometric) vector bundle over $\operatorname{Spec}\mathbb{K}$.

This construction is generalizable to free modules $M$ over a (commutative with unit) ring $R$, getting \begin{equation*} \mathscr{M}=\operatorname{Spec}R[M^{\lor}]; \end{equation*} explipictly, if $M=R^{\oplus m}$ then $\mathscr{M}\cong\mathbb{A}^m_R$ is the trivial (geometric) vector bundle ove $\operatorname{Spec}R$.

From all this, because a locally free sheaf $\mathcal{F}$ over a scheme $Y$ is locally isomorphic to some free sheaf $\widetilde{M}=\mathcal{O}_{Y|V}^{\oplus n}$ and one can prove that: \begin{equation} \mathscr{M}=\operatorname{Spec}\widetilde{M}; \end{equation} these scheme are called trivial (geometric) vector bundles $V$, and their gluing is the (geometric) vector bundle $E$ over $Y$ associated to $\mathcal{F}$.

For clarity: let $V$ be an open affine subset of $Y$ such that $\mathcal{F}_{|V}$ is isomorphic to $\mathcal{O}_{Y|V}^{\oplus n}$, in the previous notation: $R=\mathcal{O}_Y(V)$ and $\mathscr{E}_{|V}\cong\mathbb{A}^n_V=\mathbb{A}^n_R\times_{\operatorname{Spec}R}V$. Gluing on all $V$'s one constructs again $Y$; therefore, one can glue the $\mathscr{E}_{|V}$'s and constructs the total space $\mathscr{E}$ of $E$.

Let $V$ an open subset of $Y$ and let $U$ be an open subset of $X$ such that $\varphi(U)\subseteq V$; if $U$ trivializes $\varphi^{*}\mathcal{F}$ and $V$ trivializes $\mathcal{F}$, without loss of generality, $U$ and $V$ can be assume as affine and then $\varphi_{|U}$ corresponds to $\varphi_{|U}^{\sharp}:\operatorname{Spec}\mathcal{O}_Y(V)\equiv\operatorname{Spec}A\to\operatorname{Spec}\mathcal{O}_X(U)\equiv\operatorname{Spec}B$, and therefore: \begin{gather} (\varphi_{|U})^{*}\mathcal{F}_{|V}\cong(\varphi_{|U})^{*}\mathcal{O}_{Y|V}^{\oplus n}\cong(\varphi_{|U})^{*}\widetilde{A^{\oplus n}}=(\varphi_{|U})^{-1}\left(\widetilde{A^{\oplus n}}\right)\otimes_{(\varphi_{|U})^{-1}\mathcal{O}_{Y|V}}\mathcal{O}_{X|U}\cong\\ \cong\left(\varphi^{-1}\mathcal{F}\otimes_{\varphi^{-1}\mathcal{O}_Y}\mathcal{O}_X\right)_{|U}=(\varphi^{*}\mathcal{F})_{|U}=\widetilde{A^{\oplus n}\otimes_B B}=\widetilde{B^{\oplus n}}. \end{gather} By previous reasoning: \begin{equation} (\varphi^{-1}\mathscr{E})_{|U}=\operatorname{Spec}(\varphi^{*}\mathcal{F})_{|U}\cong\operatorname{Spec}(\varphi_{|U})^{*}\mathcal{F}_{|V}=(\varphi_{|U})^{-1}\mathscr{E}_{|V} \end{equation} gluing these scheme for all $U$'s and $V$'s, one constructs the (geometric) vector bundle $\varphi^{-1}E$ over $X$ associated to $\varphi^{*}\mathcal{F}$; in particular, the following diagram \begin{equation} \require{AMScd} \begin{CD} \varphi^{-1}\mathscr{E} @>>> \mathscr{E}\\ @VVV & @VVV\\ X @>>\varphi> Y \end{CD} \end{equation} is Cartesian in the category $\mathbf{Sch}$ of schemes, then $\varphi^{-1}\mathscr{E}=\mathscr{E}\times_{Y,\varphi}X$ up to canonical isomorphism.

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