Fibre product is the equalizer of a product

Suppose products and equalizers exist in $𝖢$ and we have a diagram $𝒟$

𝐴 𝑓 \to 𝐶 𝑔 \leftarrow 𝐵

Then the fibre product $l i m ⟵ 𝒟$ exists and is given by $(𝐸, 𝑝 1, 𝑝 2)$ in the commutative diagram #m/thm/cat

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where $e q$ is the equalizer of $(𝑓 𝜋 1, 𝑔 𝜋 2)$ . Conversely, any such fibre product $(𝐸, 𝑝 1, 𝑝 2)$ gives $(𝑝 1, 𝑝 2)$ as the equalizer of $(𝑓 𝜋 1, 𝑔 𝜋 2)$ .¹

Proof

Let

𝐴 𝑧 1 ⟵ 𝑍 𝑧 2 ⟶ 𝐵

such that $𝑓 𝑧 1 = 𝑔 𝑧 2$ , so $(𝑧 1, 𝑧 2) : 𝑍 \to 𝐴 \times 𝐵$ where

𝑓 𝜋 1 (𝑧 1, 𝑧 2) = 𝑔 𝜋 2 (𝑧 1, 𝑧 2) .

Now there exists $𝑢 : 𝑍 \to 𝐸$ so that $e q 𝑢 = (𝑧 1, 𝑧 2)$ . Thus

𝑝 1 𝑢 = 𝜋 1 e q 𝑢 = 𝜋 1 (𝑧 1, 𝑧 2) = 𝑧 1, 𝑝 2 𝑢 = 𝜋 2 e q 𝑢 = 𝜋 2 (𝑧 1, 𝑧 2) = 𝑧 2 .

Given an alternate $𝑢' : 𝑍 \to 𝐸$ with the property $𝑝 𝑖 𝑢' = 𝑧 𝑖$ , then $𝜋 𝑖 e q 𝑢' = 𝑧 𝑖$ so $e q 𝑢' = (𝑧 1, 𝑧 2) = e q 𝑢$ , and since the equalizer is monic $𝑢 = 𝑢'$ .

Dually, the fibre coproduct is the coëqualizer of a coproduct.

#state/tidy | #lang/en | #SemBr

2010. Category theory, ¶5.5, pp. 93–94 ↩

Fibre product is the equalizer of a product

Footnotes