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.