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Metamath Proof Explorer

Theorem fthi

Description: The morphism map of a faithful functor is an injection. (Contributed by Mario Carneiro, 27-Jan-2017)

Ref Expression
Hypotheses isfth.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
isfth.h ⊢ 𝐻 = ( Hom ‘ 𝐶 )
isfth.j ⊢ 𝐽 = ( Hom ‘ 𝐷 )
fthf1.f ⊢ ( 𝜑 → 𝐹 ( 𝐶 Faith 𝐷 ) 𝐺 )
fthf1.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
fthf1.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
fthi.r ⊢ ( 𝜑 → 𝑅 ∈ ( 𝑋 𝐻 𝑌 ) )
fthi.s ⊢ ( 𝜑 → 𝑆 ∈ ( 𝑋 𝐻 𝑌 ) )
Assertion fthi ( 𝜑 → ( ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑅 ) = ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑆 ) ↔ 𝑅 = 𝑆 ) )

Proof

Step Hyp Ref Expression
1 isfth.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
2 isfth.h ⊢ 𝐻 = ( Hom ‘ 𝐶 )
3 isfth.j ⊢ 𝐽 = ( Hom ‘ 𝐷 )
4 fthf1.f ⊢ ( 𝜑 → 𝐹 ( 𝐶 Faith 𝐷 ) 𝐺 )
5 fthf1.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
6 fthf1.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
7 fthi.r ⊢ ( 𝜑 → 𝑅 ∈ ( 𝑋 𝐻 𝑌 ) )
8 fthi.s ⊢ ( 𝜑 → 𝑆 ∈ ( 𝑋 𝐻 𝑌 ) )
9 1 2 3 4 5 6 fthf1 ⊢ ( 𝜑 → ( 𝑋 𝐺 𝑌 ) : ( 𝑋 𝐻 𝑌 ) –1-1→ ( ( 𝐹 ‘ 𝑋 ) 𝐽 ( 𝐹 ‘ 𝑌 ) ) )
10 f1fveq ⊢ ( ( ( 𝑋 𝐺 𝑌 ) : ( 𝑋 𝐻 𝑌 ) –1-1→ ( ( 𝐹 ‘ 𝑋 ) 𝐽 ( 𝐹 ‘ 𝑌 ) ) ∧ ( 𝑅 ∈ ( 𝑋 𝐻 𝑌 ) ∧ 𝑆 ∈ ( 𝑋 𝐻 𝑌 ) ) ) → ( ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑅 ) = ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑆 ) ↔ 𝑅 = 𝑆 ) )
11 9 7 8 10 syl12anc ⊢ ( 𝜑 → ( ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑅 ) = ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑆 ) ↔ 𝑅 = 𝑆 ) )