# Demystifying the Magic Diagram

Vakil calls the following pullback diagram the magic diagram. I have also seen it being called the magic square. It often shows up in fiber product diagram chases such as those associated with separatedness assertions. $$\require{AMScd}$$ $$\begin{CD} X_1\times_Y X_2 @>>> X_1\times_Z X_2\\ @V V V @VV V\\ Y @>>> Y \times_Z Y \end{CD}$$

On the other hand, I'm a firm believer that mathematics is the art of demystifying magic. So I'm interested in some intuition of this diagram. Maybe there is some geometric intuition or maybe there is a nice toy example for this?

• This explanation by Najib Idrissi might help. Mar 12, 2021 at 17:48
• @Laufen Thank you but that doesn‘t help unfortunately. I‘m perfectly aware of a proof, this question is about intuition. Mar 13, 2021 at 6:28
• Personally I find it easier to think about this diagram after transposing it: then it shows $X_1 \times_Y X_2 \to X_1 \times_Z X_2$ as the base change of the diagonal $Y \to Y \times_Z Y$ along the specified morphism $X_1 \times_Z X_2 \to Y \times_Z Y$. Mar 14, 2021 at 13:51

The diagram looks much less mysterious in the special case that $$Z$$ is the terminal object. Then the fibered products over $$Z$$ are just ordinary products, i.e., we have the diagram: $$\require{AMScd}$$ $$\begin{CD} X_1\times_Y X_2 @>>> X_1\times X_2\\ @V V V @VV V\\ Y @>>> Y \times Y \end{CD}$$ The fact that this is a pullback square can be explained intuitively as follows. Given maps $$f_1\colon X_1\to Y$$ and $$f_2\colon X_2\to Y$$, we have a natural map $$(f_1,f_2)\colon X_1\times X_2\to Y\times Y$$. Pulling back this map along the diagonal $$\Delta\colon Y\to Y\times Y$$ extracts those fibers which lie over the diagonal, which gives us the fiber product $$X_1\times_Y X_2$$.
Ok, now let's move the discussion to happen over an arbitrary object $$Z$$, i.e., in the slice category $$\mathcal{C}/Z$$. Then $$Z$$ (with the identity map) is the terminal object, and products in this category are fiber products over $$Z$$. So in the diagram above, we replace $$X_1\times X_2$$ and $$Y\times Y$$ with $$X_1\times_Z X_2$$ and $$Y\times_Z Y$$, respectively. On the other hand, fiber products in $$\mathcal{C}/Z$$ agree with fiber products in $$\mathcal{C}$$, since a square is a pullback square in $$\mathcal{C}/Z$$ if and only if it is a pullback square in $$\mathcal{C}$$ after forgetting the maps to $$Z$$. So we don't need to modify $$X_1\times_Y X_2$$ in the upper left corner. Thus we get a pullback square: $$\begin{CD} X_1\times_Y X_2 @>>> X_1\times_Z X_2\\ @V V V @VV V\\ Y @>>> Y \times_Z Y \end{CD}$$ in $$\mathcal{C}/Z$$, which (as we just noted) is also a pullback square in $$\mathcal{C}$$ after forgetting the maps to $$Z$$.
• @AndresMejia Suppose $f_1\colon X_1\to Y$ and $f_2\colon X_2\to Y$ are maps in $\mathcal{C}/Z$. Let $W$ be the pullback computed in $\mathcal{C}$, and show that $W$ has the universal property of the pullback computed in $\mathcal{C}/Z$. The key observation is that for every object in the argument, its map to $Z$ factors through $Y$, so commutativity of the square implies commutativity of all the triangles over $Z$. If you want more details, let me know! Mar 15, 2021 at 19:30