3.5. Pullback of Schemes

3.5. Pullback of Schemes === [toc] We proof in this note the following theorem : :::success **Theorem 3.5.0.1.** Pullbacks exists in the category of schemes. ::: In certain scenarios that pullbacks can be computed easily. In general situations, by properties of open immersions and techniques of gluing schemes and morphisms, one obtains the general results easily. From now on, let us fix a cospan $X\rightarrow Z\leftarrow Y$ in $\mathbf{Sch}$. 3.5.1. Simple Cases --- :::success **Lemma 3.5.1.1.** (*Pushout exists in some Special cases*) Pushout exists when either : 1. (All the objects are affine) $X,Z,Y$ are affine schemes $\operatorname{Spec}A,\operatorname{Spec}C,\operatorname{Spec}B$. 2. (One leg is an open immersion) or $X\to Z$ is an open immersion. ::: **Proof of 1.** As $\operatorname{Spec}$ is fully faithful, cospans $X\rightarrow Z\leftarrow Y$ comes from spans $A\leftarrow C\rightarrow B$. The pushout of the span $A\leftarrow C\rightarrow B$ is just $A\otimes_CB$. As $\operatorname{Spec}$ is a right adjoint functor (from $\mathbf{CRing}^{\operatorname{op}}$ to $\mathbf{sch}$), it takes pushouts (in $\mathbf{CRing}$) to pullbacks. **Proof of 2.** By using the universal property of open immersions, the pullback of $X\rightarrow Z\leftarrow Y$ is isomorphic to the open subscheme given by the inverse image of $X$ along $Z\leftarrow Y$. **Remark.** In the case of 2., **we will always consider $Y$ as the open subscheme given by the inverse image of $X$ along $Z\leftarrow Y$.** We make this choice so that we do not need to deal with additional commutative diagrams, otherwise always writing out all the explicit isomorphisms is a gruesome task when considering gluing morphisms later. 3.5.2. Lemmas concerning Global pullbacks and Local pullbacks --- We record two technical lemmas. One determines pullbacks with provided local datas, one constructs global pullbacks from local datas. This gives a nice tool for us to identify and glue pullbacks to obtain pullbacks. :::success **Lemma 3.5.2.1.** (*Detection Lemma*) Suppose $X,Y$ has open coverings $\{X_i\}_i,\{Y_j\}_j$. Given $X\leftarrow W\rightarrow Y$ completing $X\rightarrow Z\leftarrow Y$ to a commutative square, take for each $i,j$ the following diagram : $$\begin{matrix}W_{ij}&\longrightarrow&W_j&\longrightarrow&Y_j\\\downarrow&\square&\downarrow&\square&\downarrow\\W_i&\longrightarrow&W&\longrightarrow&Y\\\downarrow&\square&\downarrow&?&\downarrow\\X_i&\longrightarrow&X&\longrightarrow&Z\end{matrix}$$ where the squares decorated with a $\square$ is a pullback. If for each $i,j$ that the outer square is a pullback, the commutative square marked with a $?$ is a pullback. ::: **Sketch.** It suffices to show the case $\{X_i\}_i=\{X\}$; details for this reduction are given below. :::spoiler Assume that this case has been done. Firstly, the square given by $W_j,Y_j,X,Z$ is a pullback by applying the case to $W_{ij},W_j,Y_j,X_i,X,Z$. Therefore, the square given by $W,Y,X,Z$ is a pullback by applying the case to $W_j,W,X,Y_j,Y,Z$. ::: \ In this case, suppose $X\leftarrow V\rightarrow Y$ also completes $X\rightarrow Z\leftarrow Y$. Take the same diagram with $W$ replaced by $V$, then there are unique morphisms $V_i\to W_i$, and they agree over intersections, so they may be glued (see **Lemma 3.5.1.1.**). **Aside.** In what follows, we will only need the case $\{X_i\}_i=\{X\}$. :::success **Lemma 3.5.2.2.** (*Construction Lemma*) Suppose $X,Y$ has open coverings $\{X_i\}_i,\{Y_j\}_j$ such that $X_i\times_Z Y_j$ exists for each $i,j$, then $X\times_ZY$ exists. ::: **Sketch.** It suffices to show the case $\{X_i\}_i=\{X\}$. Write $Y_{ij}=Y_i\cap Y_j,Y_{ijk}=Y_i\cap Y_j\cap Y_k$. Let $W_i=X\times_ZY_i$. Let $W_{ij}=W_i\times_{Y_i}Y_{ij}$; this is $X\times_ZY_{ij}$. Let $W_{ijk}=W_{ij}\times_{Y_{ij}}Y_{ijk}$; this is $X\times_ZY_{ijk}$. In this sense, each $W_{ij},W_{ijk}$ are open subschemes of $W_i$ by **Lemma 3.5.1.1.**. We have unique isomorphisms $W_{ij}\to W_{ji}$ and isomorphisms $W_{ijk}\to W_{jik}$ over $W_{ij}\to W_{ji}$ and over $X,Y$. As these isomorphisms are unique, one may glue these schemes to some $W$ and also glue morphisms to obtain maps $X\leftarrow W\rightarrow Y$ completing $X\rightarrow Z\leftarrow Y$ to a commutative square. By property of gluing, we get by detection lemma that $W$ is a pullback. 3.5.3. Proof of Theorem 3.5.0.1. --- Take an open affine cover $\{Z_i\}_i$ of $Z$, define $X_i=X\times_ZZ_i$, $Y_i=Y\times_ZZ_i$. We have : $$X_i\times_ZY\simeq X_i\times_{Z_i}Y_i$$ so by **Lemma 3.5.2.2.**, may assume $Z$ is affine. Applying **Lemma 3.5.2.2.** again, may further assume $X,Y$ are affine, and we are done by **Lemma 3.5.1.1.**.