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In these two questions, it is mentioned that easy proofs of "magic square" identities can be given using the Yoneda lemma to reduce to the case of sets.

Can someone explain exactly how to do this?

In particular, how exactly to use the Yoneda lemma to prove the following lemma via the category of sets? I don't understand how to "encode these arrows as elements".

Lemma. In an arbitrary category, consider the following commutative diagram: $$\begin{matrix} X_1 & \longrightarrow & X_0 & \longleftarrow & X_2 \\\\ \downarrow & & \downarrow & & \downarrow \\\\\ S_1 & \longrightarrow & S_0 & \longleftarrow & S_2 \\\\ \uparrow & & \uparrow & & \uparrow \\\\\ Y_1 & \longrightarrow & Y_0 & \longleftarrow & Y_2 \end{matrix}$$ Assuming all the pullbacks exist, we have $$(X_1 \times_{S_1} Y_1) \times_{X_0 \times_{S_0} Y_0} (X_2 \times_{S_2} Y_2) = (X_1 \times_{X_0} X_2) \times_{S_1 \times_{S_0} S_2} (Y_1 \times_{Y_0} Y_2)$$

1 Answers1

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Roadmap:

  1. Show it in $\mathsf{Set}$.
  2. Show that if this is true in a category $\mathcal D$, then it still is true in a functor category $[\mathcal I,\mathcal D]$ for any $\mathcal I$.
  3. Remark that the Yoneda embedding $\mathcal C \hookrightarrow \widehat{\mathcal C}$ preserves all limits existing in $\mathcal C$ (in particular pullbacks).
Pece
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    Also, importantly, the Yoneda embedding reflects all limits. – Zhen Lin Dec 01 '15 at 13:46
  • @ZhenLin why is reflection of limits needed here? –  Dec 01 '15 at 15:59
  • Well, the unique functor $\mathcal{C} \to \mathbb{1}$ preserves all limits too... – Zhen Lin Dec 01 '15 at 16:15
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    @Exterior The important things is that $Y \colon \mathcal C \to \widehat{\mathcal C}$ is conservative, meaning it reflects isomorphisms (which is contained in the word embedding), so that from $Y((X_1 \times_{S_1} Y_1) \times_{X_0 \times_{S_0} Y_0} (X_2 \times_{S_2} Y_2)) \simeq Y((X_1 \times_{X_0} X_2) \times_{S_1 \times_{S_0} S_2} (Y_1 \times_{Y_0} Y_2))$ you can deduce $(X_1 \times_{S_1} Y_1) \times_{X_0 \times_{S_0} Y_0} (X_2 \times_{S_2} Y_2) \simeq (X_1 \times_{X_0} X_2) \times_{S_1 \times_{S_0} S_2} (Y_1 \times_{Y_0} Y_2)$. – Pece Dec 01 '15 at 19:15