Pushforward and pullback of morphisms

Pushforward and pullback of an isomorphism

Let 𝑓 :𝑋 →𝑌 be a morphism in an arbitrary category 𝖢. Then the following conditions are equivalent #m/thm/cat

Proof

Suppose 𝑓 :𝑋 →𝑌 is an isomorphism. Then there exists an inverse 𝑔 =𝑓−1 :𝑌 →𝑋. For any 𝑍 ∈𝖢, there exist pushforwards 𝑓⋆ :𝖢(𝑍,𝑋) →𝖢(𝑍,𝑌) and 𝑔⋆ :𝖢(𝑍,𝑌) →𝖢(𝑍,𝑋). Let 𝑠 :𝑍 →𝑋, then clearly 𝑔⋆𝑓⋆(𝑠) =𝑔⋆(𝑓𝑠) =𝑔𝑓𝑠 =𝑠. Likewise for 𝑡 :𝑌 →𝑋, clearly 𝑓⋆𝑔⋆(𝑡) =𝑓⋆(𝑔𝑡) =𝑓𝑔𝑡 =𝑡. Hence 𝑔⋆ is the inverse of 𝑓⋆ Similarly for any 𝑍 ∈𝖢, there exist pullbacks 𝑓⋆ :𝖢(𝑌,𝑍) →𝖢(𝑋,𝑍) and 𝑔⋆ :𝖢(𝑋,𝑍) →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 ∈𝖢 the pushforward 𝑓⋆ :𝖢(𝑍,𝑋) →𝖢(𝑍,𝑌) is a bijection. If we let 𝑍 =𝑌, from surjectivity it follows there exists 𝑔 :𝑌 →𝑋 such that 𝑓⋆(𝑔) =𝑓𝑔 =id𝑌. If we let 𝑍 =𝑋, it follows that 𝑓⋆(𝑔𝑓) =𝑓𝑔𝑓 =𝑓 =𝑓⋆(id𝑋), and hence from injectivity 𝑔𝑓 =id𝑋. Therefore 𝑔 is the inverse of 𝑓, whence 𝑓 is an isomorphism.

Finally, assume for any 𝑍 ∈𝖢 the pullback 𝑓⋆ :𝖢(𝑌,𝑍) →𝖢(𝑋,𝑍) is a bijection. If we let 𝑍 =𝑋, from surjectivity it follows there exists 𝑔 :𝑌 →𝑋 such that 𝑓⋆(𝑔) =𝑔𝑓 =id𝑋. If we let 𝑍 =𝑌, it follows that 𝑓⋆(𝑓𝑔) =𝑓𝑔𝑓 =𝑔 =𝑓⋆(id𝑌), and hence from injectivity 𝑓𝑔 =id𝑌. Therefore 𝑔 is the inverse of 𝑓, whence 𝑓 is an isomorphism.

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


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

Footnotes

  1. 2020, Topology: A categorical approach, p. 9 ↩