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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?

Alex Kruckman
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Qi Zhu
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    This explanation by Najib Idrissi might help. – Jessie Mar 12 '21 at 17:48
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    @Laufen Thank you but that doesn‘t help unfortunately. I‘m perfectly aware of a proof, this question is about intuition. – Qi Zhu Mar 13 '21 at 06:28
  • 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 $X_1 \times_Z X_2 \to Y \times_Z Y$. – Zhen Lin Mar 14 '21 at 13:51

2 Answers2

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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 is easy to check and 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 $(f_1\times f_2)\colon X_1\times X_2\to Y\times Y$. 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 $Z$.

Alex Kruckman
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    this is brilliant. Is there an easy see to see why that forgetful map preserves pullbacks? – Andres Mejia Mar 15 '21 at 18:15
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    @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! – Alex Kruckman Mar 15 '21 at 19:30
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    Actually that was fair enough. I was thinking it might be more “right adjoint” or some nonsense like that. This is great, and seems like a proof to me – Andres Mejia Mar 15 '21 at 19:31
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    @AndresMejia Yeah, normally you'd want to show that a functor preserves pullbacks by showing that it's a right adjoint. But the forgetful functor here fails to preserve other limits, e.g. products and the terminal object. I'd be interested to know if there's a "fancy" reason why this functor preserves pullbacks and equalizers. – Alex Kruckman Mar 15 '21 at 19:35
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    It actually preserves limits of all connected diagrams, iirc. – Zhen Lin Mar 23 '21 at 11:33
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Here is some geometric intuition. If we are working with sets (think fiber bundle), one way to view the pullback of $E \to B$ via $f \colon B' \to B$ is as the union $$ f^*B' \cong \bigsqcup_{b' \in B'} E_{f(b')} .$$ In other words, the fiber of $f(b')$ gets pulled back to form the fiber of $b'$. Alternatively, we may take a more symmetric view, and write the fiber product of $E$ and $F$ over $B$ as $$ E \times_{B} F \cong \bigsqcup_{b \in B} E_b \times E_f. $$ With these two viewpoints, we can better understand the geometry of the magic diagram.

Since we have a sequence of maps $X_i \to Y \to Z$, we see that fibers of $X_i$ over $Z$ map onto fibers of $Y$ over $Z$. Combining the data of all these maps, we obtain a map $X_1 \times_Z X_2 \to Y \times_Z Y$, whose fibers are products of fibers of the maps $X_i \to Y$.

For each fiber $Y_z$ of $Y$ over $Z$, we obtain a diagonal map $\delta_z \colon Y_z \to Y_z \times Y_z$, which may be collated to form a single diagonal map $\delta \colon Y \to Y \times_Z Y$. It follows that

$$ \delta^* (X_1 \times_Z X_2) = \bigsqcup_{y \in Y} (X_1 \times_Z X_2)_{(y, y)} = \bigsqcup_{y \in Y} (X_1)_y \times (X_2)_y. $$

In other words, the pullback along the diagonal map is the disjoint union of the product of the fibers of $X_1$ and $X_2$ over a common point in $Y$. But this is just the fiber product of $X_1$ and $X_2$ over $Y$.

Finally, a bit non-rigorous discussion about what this means. We can describe the situation as having bundle maps $X_i \to Z$ which factor into composition of bundle maps $X_i \to Y \to Z$. If $Z$ is the base space and $X_i$ is the total space, then we may think of $Y$ as a sort of "partial" space. Similarly, we can think of fibers of $X_i$ over $Z$ as being "total" fibers and fibers of $X_i$ over $Y$ as being "partial" fibers. By "stopping early" and considering maps from $X_i$ to $Y$ instead of $X_i$ to $Z$, we can increase the "granularity" of the bundle, turning total fibers into partial fibers.