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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?

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    $\begingroup$ This explanation by Najib Idrissi might help. $\endgroup$
    – Jessie
    Mar 12, 2021 at 17:48
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    $\begingroup$ @Laufen Thank you but that doesn‘t help unfortunately. I‘m perfectly aware of a proof, this question is about intuition. $\endgroup$
    – Qi Zhu
    Mar 13, 2021 at 6:28
  • $\begingroup$ 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 $X_1 \times_Z X_2 \to Y \times_Z Y$. $\endgroup$
    – Zhen Lin
    Mar 14, 2021 at 13:51

1 Answer 1

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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 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 $Z$.

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    $\begingroup$ this is brilliant. Is there an easy see to see why that forgetful map preserves pullbacks? $\endgroup$ Mar 15, 2021 at 18:15
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    $\begingroup$ @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! $\endgroup$ Mar 15, 2021 at 19:30
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    $\begingroup$ 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 $\endgroup$ Mar 15, 2021 at 19:31
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    $\begingroup$ @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. $\endgroup$ Mar 15, 2021 at 19:35
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    $\begingroup$ It actually preserves limits of all connected diagrams, iirc. $\endgroup$
    – Zhen Lin
    Mar 23, 2021 at 11:33

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