Does base change preserve fibre products?

Question

Let $T_1, T_2, T$ be schemes over a base scheme $X$ ,i.e. they are $X$-schemes. Now, let $f:T_1 \rightarrow T$ and $f:T_1 \rightarrow T$ be two $X$-scheme morphism. So, we can compute the fibre product the $T_1 \times_T T_2 $.

Now, let $U$ be another $X$-scheme. We can compute the base change, $T_1 \times_X U$, $T_2 \times_X U$ and $T \times_X U$ and gain compute the fibre product $(T_1 \times_X U) \times_{(T \times_X U)} (T_2 \times_X U)$.

Is $(T_1 \times_X U) \times_{(T \times_X U)} (T_2 \times_X U)$ isomorphic to $(T_1 \times_T T_2) \times_X U $ ?

There exists a canonical morphism from $(T_1 \times_T T_2) \times_X U$ to $(T_1 \times_X U) \times_{(T \times_X U)} (T_2 \times_X U)$. I think this is an isomorphic but I'm not completely sure (there may be a mistake in my proof)

Answer 1

Yes. This is a statement about products and pullbacks in the category of schemes over $X$; I will prove it in an arbitrary category. The easiest way to show it is to show that these two objects have the same universal property. Let us write $$V=(T_1 \times_T T_2) \times U$$ and $$W=(T_1 \times U) \times_{T \times U} (T_2 \times U).$$ Here $\times$ with no subscript is the product in our category, which is the fiber product $\times_X$ in the case of the category of schemes over $X$.

For any object $A$, maps $A\to V$ are naturally in bijection with pairs of maps $g:A\to T_1\times_T T_2$ and $hg:A\to U$. Such maps $f$ are in turn naturally in bijection with pairs of maps $g_1:A\to T_1$ and $g_2:A\to T_2$ such that $f_1g_1=f_2g_2$, where $f_1:T_1\to T$ and $f_2:T_2\to T$ are the given maps. So maps $A\to V$ are naturally in bijection with triples $(g_1,g_2,h)$ with $g_1:A\to T_1$, $g_2:A\to T_2$, $h:A\to U$, and $f_1g_1=f_2g_2$.

On the other hand, maps $A\to W$ are naturally in bijection with pairs of maps $i_1:A\to T_1\times U$ and $i_2:A\to T_2\times U$ such that $(f_1\times 1_U)i_1=(f_2\times 1_U)i_2$. The map $i_1$ is uniquely determined by a pair of maps $g_1:A\to T_1$ and $h_1:A\to U$, and similarly $i_2$ is determined by a pair of maps $g_2:A\to T_2$ and $h_2:A\to U$. Note then that $$(f_1\times 1_U)i_1=(f_1\times 1_U)(g_1,h_1)=(f_1g_1,1_Uh_1)=(f_1g_1,h_1)$$ and similarly $(f_2\times 1_U)i_2=(f_2g_2,h_2)$. So the condition $(f_1\times 1_U)i_1=(f_2\times 1_U)i_2$ is equivalent to requiring $f_1g_1=f_2g_2$ and $h_1=h_2$. Writing $h$ for the common value of $h_1$ and $h_2$, we thus see that maps $A\to W$ are naturally in bijection with triples $(g_1,g_2,h)$ with $g_1:A\to T_1$, $g_2:A\to T_2$, $h:A\to U$, and $f_1g_1=f_2g_2$.

Thus for any object $A$, maps $A\to V$ and $A\to W$ are naturally in bijection with each other. By Yoneda, this implies $V\cong W$.

Answer 2

Here's another perspective. Like Eric Wofsey, I'll consider an arbitrary category, $\mathcal C$, with (chosen) fibered products (aka pullbacks). For each object $X$ of $\mathcal C$ we can consider the slice category $\mathcal C/X$ whose objects are arrows into $X$ and arrows are commuting triangles. Given any arrow $f:U\to X$, we have the change of base functor $f^* : \mathcal C/X\to\mathcal C/U$. This takes an arrow $p:T\to X$ to the projection to $U$ of $U\times_{f,p}T$ as a diagram: $$\require{AMScd}\begin{CD} U\times_{f,p}T @>>> T \\ @Vf^*(p)VV @VVpV \\ U @>>f> X\end{CD}$$

Your question is then whether $f^*$ preserves pullbacks. Since pullbacks are limits, this would be true if $f^*$ preserved all limits which it would if it was a right adjoint. Conveniently, $f^*$ is, in fact, a right adjoint.

We need to show that there is a functor $f_! : \mathcal C/U\to\mathcal C/X$ such that an arrow (over $X$) $g:f_!(p)\to q$, i.e. $$\begin{CD} \mathsf{dom}(f_!(p)) @>g>> B\\ @Vf_!(p)VV @VVqV \\ X @= X\end{CD}$$ is the same thing as an arrow $p\to f^*(q)$, i.e. $$\begin{CD} A @>\overline g>> U\times_{f,q}B\\ @VpVV @VVf^*(q)V \\ U @= U\end{CD}$$ There's an obvious way of turning an arrow $A\to U$ into an arrow $A\to X$ given an arrow $f:U\to X$. Just post-compose with $f$! The first diagram then looks like:$$\begin{CD} A @>g>> B\\ @VpVV @VVqV \\ U @>>f> X\end{CD}$$ i.e. $f\circ p = q\circ g$ which, via the universal property of pullbacks, gives us a unique arrow $A\to U\times_{f,q}B$ for which gives $p$ when post-composed with $f^*(q)$, i.e. is an arrow $p\to f^*(q)$.

Conversely, given an arrow $\overline g:A\to U\times_{f,q}B$ such that $p=f^*(q)\circ\overline g$, we can make an arrow $A\to B$ simply by using the other projection of the fibered product giving $A\stackrel{\overline g}{\to}U\times_{f,q}B\to B$ which gives an arrow $f_!(p)\to q$ via the outer rectangle of $$\begin{CD} A @>\overline g>> U\times_{f,q} B @>>> B \\ @VpVV @Vf^*(q)VV @VVqV \\ U @= U @>>f> X \end{CD}$$

It remains to show that this bijection is natural in $p$ and $q$ to get $f_!\dashv f^*$, but I'll leave that to you.