Metamath Proof Explorer


Theorem coffth

Description: The composition of two fully faithful functors is fully faithful. (Contributed by Mario Carneiro, 28-Jan-2017)

Ref Expression
Hypotheses coffth.f ( 𝜑𝐹 ∈ ( ( 𝐶 Full 𝐷 ) ∩ ( 𝐶 Faith 𝐷 ) ) )
coffth.g ( 𝜑𝐺 ∈ ( ( 𝐷 Full 𝐸 ) ∩ ( 𝐷 Faith 𝐸 ) ) )
Assertion coffth ( 𝜑 → ( 𝐺func 𝐹 ) ∈ ( ( 𝐶 Full 𝐸 ) ∩ ( 𝐶 Faith 𝐸 ) ) )

Proof

Step Hyp Ref Expression
1 coffth.f ( 𝜑𝐹 ∈ ( ( 𝐶 Full 𝐷 ) ∩ ( 𝐶 Faith 𝐷 ) ) )
2 coffth.g ( 𝜑𝐺 ∈ ( ( 𝐷 Full 𝐸 ) ∩ ( 𝐷 Faith 𝐸 ) ) )
3 1 elin1d ( 𝜑𝐹 ∈ ( 𝐶 Full 𝐷 ) )
4 2 elin1d ( 𝜑𝐺 ∈ ( 𝐷 Full 𝐸 ) )
5 3 4 cofull ( 𝜑 → ( 𝐺func 𝐹 ) ∈ ( 𝐶 Full 𝐸 ) )
6 1 elin2d ( 𝜑𝐹 ∈ ( 𝐶 Faith 𝐷 ) )
7 2 elin2d ( 𝜑𝐺 ∈ ( 𝐷 Faith 𝐸 ) )
8 6 7 cofth ( 𝜑 → ( 𝐺func 𝐹 ) ∈ ( 𝐶 Faith 𝐸 ) )
9 5 8 elind ( 𝜑 → ( 𝐺func 𝐹 ) ∈ ( ( 𝐶 Full 𝐸 ) ∩ ( 𝐶 Faith 𝐸 ) ) )