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 ( 𝜑𝐹 ( 𝐶 Faith 𝐷 ) 𝐺 )
fthf1.x ( 𝜑𝑋𝐵 )
fthf1.y ( 𝜑𝑌𝐵 )
fthi.r ( 𝜑𝑅 ∈ ( 𝑋 𝐻 𝑌 ) )
fthi.s ( 𝜑𝑆 ∈ ( 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 ( 𝜑𝐹 ( 𝐶 Faith 𝐷 ) 𝐺 )
5 fthf1.x ( 𝜑𝑋𝐵 )
6 fthf1.y ( 𝜑𝑌𝐵 )
7 fthi.r ( 𝜑𝑅 ∈ ( 𝑋 𝐻 𝑌 ) )
8 fthi.s ( 𝜑𝑆 ∈ ( 𝑋 𝐻 𝑌 ) )
9 1 2 3 4 5 6 fthf1 ( 𝜑 → ( 𝑋 𝐺 𝑌 ) : ( 𝑋 𝐻 𝑌 ) –1-1→ ( ( 𝐹𝑋 ) 𝐽 ( 𝐹𝑌 ) ) )
10 f1fveq ( ( ( 𝑋 𝐺 𝑌 ) : ( 𝑋 𝐻 𝑌 ) –1-1→ ( ( 𝐹𝑋 ) 𝐽 ( 𝐹𝑌 ) ) ∧ ( 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 ) )