Question

5.5.7. Let α A → B and β : B → C be invertible mappings. (a) Prove that β ◦α : A → C is invertible.

Answer 1

If \(\alpha :A\to B,\beta :B\to C\) are invertible then there exist maps \(\alpha^{-1}:B\to A,\beta^{-1}:C\to B\) such that \(\alpha\circ\alpha^{-1}=\alpha^{-1}\circ\alpha=\operatorname{Id}\) and likewise for \(\beta,\beta^{-1}\)

Answer 2

I should clarify: \(\operatorname{Id}_A\) for \(\alpha\) and then \(\operatorname{Id}_B\) for \(\beta\)

Answer 3

Now consider the map \(\gamma:A\to C\) defined as \(\gamma=\beta\circ\alpha\). Since we know \(\alpha,\beta\) are invertible it's clear that \((\alpha^{-1}\circ\beta^{-1})\circ\gamma=\alpha^{-1}\circ(\beta^{-1}\circ(\beta\circ\alpha))=\alpha^{-1}\circ((\beta^{-1}\circ\beta)\circ\alpha)=\alpha^{-1}\circ\alpha=\operatorname{Id}_A\). The other case follows similarly. Clearly \(\gamma^{-1}=\alpha^{-1}\circ\beta^{-1}\) fits the definition for invertibility.