Altanis

1.7Exercises

Updated 11 May 2026Chapter (PDF)
[1.7.1]Problem#

Let A,B∈Obj(C)A, B \in \Obj(\sf{C}) with f∈Hom⁡C(A,B)f \in \Hom_{\sf{C}}(A, B).

  1. Show that if ff has a right inverse g∈Hom⁡C(B,A)g \in \Hom_{\sf{C}}(B, A), then ff is an epimorphism.

  2. Show that the converse is not true by explicitly constructing a category and an epimorphism without a right-inverse.

Solution.

Let β′,β′′∈Hom⁡C(C,A)\beta', \beta'' \in \Hom_{\sf{C}}(C, A), with C∈Obj(C)C \in \Obj(\sf{C}), be such that

β′∘f=β′′∘f.\beta' \circ f = \beta'' \circ f.

Apply gg to both sides, yielding

β′∘(f∘g)=β′′∘(f∘g)  ⟹  β′∘1B=β′′∘1B  ⟹  β′=β′′,\beta' \circ (f \circ g) = \beta'' \circ (f \circ g) \implies \beta' \circ 1_B = \beta'' \circ 1_B \implies \beta' = \beta'',

completing (1)(1).

To do (2)(2), consider the category C\sf{C} induced by the partial ordering ≤\le on Z\bZ. Then fix the morphism f:2↠3f: 2 \twoheadrightarrow 3, which is right-cancellative by inspection (and thus epic). However note ff has no right inverse (33 can never map to 22), completing the proof.