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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 ⊢ ( 𝜑 → 𝐶 ∈ Cat )
isepi.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
isepi.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
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 ⊢ ( 𝜑 → 𝐶 ∈ Cat )
6 isepi.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
7 isepi.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
8 1 2 3 4 5 6 7 isepi ⊢ ( 𝜑 → ( 𝑓 ∈ ( 𝑋 𝐸 𝑌 ) ↔ ( 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ∧ ∀ 𝑧 ∈ 𝐵 Fun ◡ ( 𝑔 ∈ ( 𝑌 𝐻 𝑧 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌 ⟩ · 𝑧 ) 𝑓 ) ) ) ) )
9 simpl ⊢ ( ( 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ∧ ∀ 𝑧 ∈ 𝐵 Fun ◡ ( 𝑔 ∈ ( 𝑌 𝐻 𝑧 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌 ⟩ · 𝑧 ) 𝑓 ) ) ) → 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) )
10 8 9 biimtrdi ⊢ ( 𝜑 → ( 𝑓 ∈ ( 𝑋 𝐸 𝑌 ) → 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ) )
11 10 ssrdv ⊢ ( 𝜑 → ( 𝑋 𝐸 𝑌 ) ⊆ ( 𝑋 𝐻 𝑌 ) )