Metamath Proof Explorer


Theorem epihom

Description: An epimorphism is a morphism. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses isepi.b 𝐵 = ( Base ‘ 𝐶 )
isepi.h 𝐻 = ( Hom ‘ 𝐶 )
isepi.o · = ( comp ‘ 𝐶 )
isepi.e 𝐸 = ( Epi ‘ 𝐶 )
isepi.c ( 𝜑𝐶 ∈ Cat )
isepi.x ( 𝜑𝑋𝐵 )
isepi.y ( 𝜑𝑌𝐵 )
Assertion epihom ( 𝜑 → ( 𝑋 𝐸 𝑌 ) ⊆ ( 𝑋 𝐻 𝑌 ) )

Proof

Step Hyp Ref Expression
1 isepi.b 𝐵 = ( Base ‘ 𝐶 )
2 isepi.h 𝐻 = ( Hom ‘ 𝐶 )
3 isepi.o · = ( comp ‘ 𝐶 )
4 isepi.e 𝐸 = ( Epi ‘ 𝐶 )
5 isepi.c ( 𝜑𝐶 ∈ Cat )
6 isepi.x ( 𝜑𝑋𝐵 )
7 isepi.y ( 𝜑𝑌𝐵 )
8 1 2 3 4 5 6 7 isepi ( 𝜑 → ( 𝑓 ∈ ( 𝑋 𝐸 𝑌 ) ↔ ( 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ∧ ∀ 𝑧𝐵 Fun ( 𝑔 ∈ ( 𝑌 𝐻 𝑧 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌· 𝑧 ) 𝑓 ) ) ) ) )
9 simpl ( ( 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ∧ ∀ 𝑧𝐵 Fun ( 𝑔 ∈ ( 𝑌 𝐻 𝑧 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌· 𝑧 ) 𝑓 ) ) ) → 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) )
10 8 9 syl6bi ( 𝜑 → ( 𝑓 ∈ ( 𝑋 𝐸 𝑌 ) → 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ) )
11 10 ssrdv ( 𝜑 → ( 𝑋 𝐸 𝑌 ) ⊆ ( 𝑋 𝐻 𝑌 ) )