Metamath Proof Explorer


Theorem comffn

Description: The functionalized composition operation is a function. (Contributed by Mario Carneiro, 4-Jan-2017)

Ref Expression
Hypotheses comfffn.o 𝑂 = ( compf𝐶 )
comfffn.b 𝐵 = ( Base ‘ 𝐶 )
comffn.h 𝐻 = ( Hom ‘ 𝐶 )
comffn.x ( 𝜑𝑋𝐵 )
comffn.y ( 𝜑𝑌𝐵 )
comffn.z ( 𝜑𝑍𝐵 )
Assertion comffn ( 𝜑 → ( ⟨ 𝑋 , 𝑌𝑂 𝑍 ) Fn ( ( 𝑌 𝐻 𝑍 ) × ( 𝑋 𝐻 𝑌 ) ) )

Proof

Step Hyp Ref Expression
1 comfffn.o 𝑂 = ( compf𝐶 )
2 comfffn.b 𝐵 = ( Base ‘ 𝐶 )
3 comffn.h 𝐻 = ( Hom ‘ 𝐶 )
4 comffn.x ( 𝜑𝑋𝐵 )
5 comffn.y ( 𝜑𝑌𝐵 )
6 comffn.z ( 𝜑𝑍𝐵 )
7 eqid ( 𝑔 ∈ ( 𝑌 𝐻 𝑍 ) , 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑍 ) 𝑓 ) ) = ( 𝑔 ∈ ( 𝑌 𝐻 𝑍 ) , 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑍 ) 𝑓 ) )
8 ovex ( 𝑔 ( ⟨ 𝑋 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑍 ) 𝑓 ) ∈ V
9 7 8 fnmpoi ( 𝑔 ∈ ( 𝑌 𝐻 𝑍 ) , 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑍 ) 𝑓 ) ) Fn ( ( 𝑌 𝐻 𝑍 ) × ( 𝑋 𝐻 𝑌 ) )
10 eqid ( comp ‘ 𝐶 ) = ( comp ‘ 𝐶 )
11 1 2 3 10 4 5 6 comffval ( 𝜑 → ( ⟨ 𝑋 , 𝑌𝑂 𝑍 ) = ( 𝑔 ∈ ( 𝑌 𝐻 𝑍 ) , 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑍 ) 𝑓 ) ) )
12 11 fneq1d ( 𝜑 → ( ( ⟨ 𝑋 , 𝑌𝑂 𝑍 ) Fn ( ( 𝑌 𝐻 𝑍 ) × ( 𝑋 𝐻 𝑌 ) ) ↔ ( 𝑔 ∈ ( 𝑌 𝐻 𝑍 ) , 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) ↦ ( 𝑔 ( ⟨ 𝑋 , 𝑌 ⟩ ( comp ‘ 𝐶 ) 𝑍 ) 𝑓 ) ) Fn ( ( 𝑌 𝐻 𝑍 ) × ( 𝑋 𝐻 𝑌 ) ) ) )
13 9 12 mpbiri ( 𝜑 → ( ⟨ 𝑋 , 𝑌𝑂 𝑍 ) Fn ( ( 𝑌 𝐻 𝑍 ) × ( 𝑋 𝐻 𝑌 ) ) )