Pushforward and pullback of an isomorphism

Let $𝑓 : 𝑋 \to 𝑌$ be a morphism in an arbitrary category $𝖢$ , and $𝑋, 𝑌, 𝑍 \in 𝖢$ Then the following conditions are equivalent cat

$𝑓 : 𝑋 \to 𝑌$ is an isomorphism
$𝑓 ⋆ : 𝖢 (𝑍, 𝑋) \to 𝖢 (𝑍, 𝑌)$ is a bijection
$𝑓 ⋆ : 𝖢 (𝑌, 𝑍) \to 𝖢 (𝑋, 𝑍)$ is a bijection

Proof

Suppose $𝑓 : 𝑋 \to 𝑌$ is an isomorphism. Then there exists an inverse $𝑔 = 𝑓 - 1 : 𝑌 \to 𝑋$ . For any $𝑍 \in 𝖢$ , there exist pushforwards $𝑓 ⋆ : 𝖢 (𝑍, 𝑋) \to 𝖢 (𝑍, 𝑌)$ and $𝑔 ⋆ : 𝖢 (𝑍, 𝑌) \to 𝖢 (𝑍, 𝑋)$ . Let $𝑠 : 𝑍 \to 𝑋$ , then clearly $𝑔 ⋆ 𝑓 ⋆ (𝑠) = 𝑔 ⋆ (𝑓 𝑠) = 𝑔 𝑓 𝑠 = 𝑠$ . Likewise for $𝑡 : 𝑌 \to 𝑋$ , clearly $𝑓 ⋆ 𝑔 ⋆ (𝑡) = 𝑓 ⋆ (𝑔 𝑡) = 𝑓 𝑔 𝑡 = 𝑡$ . Hence $𝑔 ⋆$ is the inverse of $𝑓 ⋆$ Similarly for any $𝑍 \in 𝖢$ , there exist pullbacks $𝑓 ⋆ : 𝖢 (𝑌, 𝑍) \to 𝖢 (𝑋, 𝑍)$ and $𝑔 ⋆ : 𝖢 (𝑋, 𝑍) \to 𝖢 (𝑌, 𝑍)$ . Proceeding as before, $𝑔 ⋆$ is the inverse of $𝑓 ⋆$ . Therefore, if $𝑓$ is an isomorphism, so are $𝑓 ⋆$ and $𝑓 ⋆$ bijections.

Next, assume for any $𝑍 \in 𝖢$ the pushforward $𝑓 ⋆ : 𝖢 (𝑍, 𝑋) \to 𝖢 (𝑍, 𝑌)$ is a bijection. If we let $𝑍 = 𝑌$ , from surjectivity it follows there exists $𝑔 : 𝑌 \to 𝑋$ such that $𝑓 ⋆ (𝑔) = 𝑓 𝑔 = i d 𝑌$ . If we let $𝑍 = 𝑋$ , it follows that $𝑓 ⋆ (𝑔 𝑓) = 𝑓 𝑔 𝑓 = 𝑓 = 𝑓 ⋆ (i d 𝑋)$ , and hence from injectivity $𝑔 𝑓 = i d 𝑋$ . Therefore $𝑔$ is the inverse of $𝑓$ , whence $𝑓$ is an isomorphism.

Finally, assume for any $𝑍 \in 𝖢$ the pullback $𝑓 ⋆ : 𝖢 (𝑌, 𝑍) \to 𝖢 (𝑋, 𝑍)$ is a bijection. If we let $𝑍 = 𝑋$ , from surjectivity it follows there exists $𝑔 : 𝑌 \to 𝑋$ such that $𝑓 ⋆ (𝑔) = 𝑔 𝑓 = i d 𝑋$ . If we let $𝑍 = 𝑌$ , it follows that $𝑓 ⋆ (𝑓 𝑔) = 𝑓 𝑔 𝑓 = 𝑔 = 𝑓 ⋆ (i d 𝑌)$ , and hence from injectivity $𝑓 𝑔 = i d 𝑌$ . Therefore $𝑔$ is the inverse of $𝑓$ , whence $𝑓$ is an isomorphism.

In summary, if you understand all the morphisms $𝑋 \to 𝑍$ , you know $𝑋$ up to isomorphism. In summary, if you understand all the morphisms $𝑍 \to 𝑋$ , you know $𝑋$ up to isomorphism.¹

tidy | en | SemBr

2020, Topology: A categorical approach, p. 9 ↩

Pushforward and pullback of an isomorphism

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