Where does this attempted proof of the "magic diagram" being Cartesian fail?

Question

I'm trying to prove the Diagonal-Base-Change-Diagram (i.e. "magic diagram") from Vakil's book, and I can't find a mistake in my proof attempt, even though it's clearly incorrect:

We need to prove, given morphisms $X_1, X_2 \rightarrow Y$ and $Y \rightarrow Z$, that the following diagram is a Cartesian square (i.e. fibered product):

$$ \begin{array}{ccc} X_1\times_Y X_2 & \longrightarrow & X_1\times_Z X_2 \\ \Big\downarrow && \Big\downarrow \\ Y & \longrightarrow & Y\times_Z Y \end{array} $$

I first expanded out into the following diagram on the right:

The subdiagrams clearly commute, because:

• Square 5 commutes by the definition of $X_1 \times_Z X_2$
• Square 3 and 4 commutes, as it's defining the $X_1 \rightarrow Z$ and $X_2 \rightarrow Z$ maps to be the composition of the maps $\alpha: X_1 \rightarrow Y$, $\beta: X_2 \rightarrow Y$ and $\gamma: Y \rightarrow Z$ provided by the problem.
• Square 1 and 2 commutes by the universal property of $X_1 \times_Z X_2$: the $X_1 \times_Y X_2$ is a fibered product over $\alpha: X_1 \rightarrow Y$ and $X_2 \rightarrow Y$ provided by the problem; so by the definition of a fibered product, we have the projection maps $X_1 \times_Y X_2 \rightarrow X_1$ and $X_1 \times_Y X_2 \rightarrow X_2$. And composing each with the $X_1 \rightarrow Z$ and $X_2 \rightarrow Z$ map, the two resulting $X_1 \times_Y X_2 \rightarrow Z$ maps commute; therefore, by the universal property of $X_1 \times_Z X_2$, we have a unique map $X_1 \times_Y X_2 \rightarrow X_1 \times_Z X_2$.
• The stuff at the bottom-right below the rectangle involving $Y$ and $Y \times_Z Y$ are simply the definition of the diagonal map.

And because pasting of commutative diagrams are commutative, so is the entire diagram in the image provided. Then, I attempted to show that the diagram provided by the problem is indeed a fibered product by considering some object $O$, and morphisms $O \rightarrow X_1 \times_Z X_2$ and $O \rightarrow Y$.

First considering $O \rightarrow X_1 \times_Z X_2$, compose it with the projection map to get $O \rightarrow X_1$ and $O \rightarrow X_2$, and compose them each with $\alpha$ and $\beta$ to get two $O \rightarrow Y$ morphisms. Finally, since the diagram is commutative, these two morphisms must be identical.

But this is clearly incorrect, as we never used the $O \rightarrow Y$ map.

I am aware of correct proofs of this theorem (such as this previous question), but even after reading those I can't find which step of my attempted proof is incorrect.

Answer 1

Nice question! Hopefully my answer makes things clearer

Concretely here, (thinking about sets for instance) an element of $X_1\times_Z X_2$ is a pair $(x_1,x_2)$ such that $\gamma(\alpha(x_1))=\gamma(\beta(x_2))$ in $Z$, but it could happen that $\alpha(x_1)\neq \beta(x_2)$ in $Y$!

If your original diagram was commutative as you use it, then you could erase $X_1 \times_Y X_2$ (along with the arrows coming out of it) and still find a commutative diagram, which may not be the case by the argument above.

Here is what goes wrong. This is the commutative diagram you have before introducing any fiber product: $\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\la}{\kern-1.5ex\longleftarrow \phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} X_1 & \ra{} & Y & \la & X_2 \\ \da{id} & & \da{} & & \da{id} \\ X_1 & \ras{} & Z & \la & X_2 \\ \end{array}$

You can clearly see here that $Y$ is not a target in a square, which is what is causing the issue. Being a commutative diagram precisely means that the different paths from "source" to "target" agree, where it is important to understand who plays the role of the source, of the target, or else of an intermediate object! If two such paths happen to go through some common intermediate object halfway through, that doesn't mean they already agree here!

If you add $O$ above, mapping to $X_1 \times_Z X_2$, then to $X_1$ and $X_2$, the diagram only commutes if you go all the way to $Z$ (by definition of $X_1 \times_Z X_2$), not necessarily if you stop at $Y$. To make this even clearer, I could have split $Y$ into two separate instances, each mapping to the same $Z$.

The fact that $O$ also maps to $Y$, and that both maps from $O$ to $Y \times_Z Y$ agree, ensures that the arrows already agree at $Y$, and proves the claim.