Metamath Proof Explorer


Theorem diag1a

Description: The constant functor of X . (Contributed by Zhi Wang, 19-Oct-2025)

Ref Expression
Hypotheses diag1.l 𝐿 = ( 𝐶 Δfunc 𝐷 )
diag1.c ( 𝜑𝐶 ∈ Cat )
diag1.d ( 𝜑𝐷 ∈ Cat )
diag1.a 𝐴 = ( Base ‘ 𝐶 )
diag1.x ( 𝜑𝑋𝐴 )
diag1.k 𝐾 = ( ( 1st𝐿 ) ‘ 𝑋 )
diag1.b 𝐵 = ( Base ‘ 𝐷 )
diag1.j 𝐽 = ( Hom ‘ 𝐷 )
diag1.i 1 = ( Id ‘ 𝐶 )
Assertion diag1a ( 𝜑𝐾 = ⟨ ( 𝐵 × { 𝑋 } ) , ( 𝑦𝐵 , 𝑧𝐵 ↦ ( ( 𝑦 𝐽 𝑧 ) × { ( 1𝑋 ) } ) ) ⟩ )

Proof

Step Hyp Ref Expression
1 diag1.l 𝐿 = ( 𝐶 Δfunc 𝐷 )
2 diag1.c ( 𝜑𝐶 ∈ Cat )
3 diag1.d ( 𝜑𝐷 ∈ Cat )
4 diag1.a 𝐴 = ( Base ‘ 𝐶 )
5 diag1.x ( 𝜑𝑋𝐴 )
6 diag1.k 𝐾 = ( ( 1st𝐿 ) ‘ 𝑋 )
7 diag1.b 𝐵 = ( Base ‘ 𝐷 )
8 diag1.j 𝐽 = ( Hom ‘ 𝐷 )
9 diag1.i 1 = ( Id ‘ 𝐶 )
10 1 2 3 4 5 6 7 8 9 diag1 ( 𝜑𝐾 = ⟨ ( 𝑦𝐵𝑋 ) , ( 𝑦𝐵 , 𝑧𝐵 ↦ ( 𝑓 ∈ ( 𝑦 𝐽 𝑧 ) ↦ ( 1𝑋 ) ) ) ⟩ )
11 fconstmpt ( 𝐵 × { 𝑋 } ) = ( 𝑦𝐵𝑋 )
12 fconstmpt ( ( 𝑦 𝐽 𝑧 ) × { ( 1𝑋 ) } ) = ( 𝑓 ∈ ( 𝑦 𝐽 𝑧 ) ↦ ( 1𝑋 ) )
13 12 a1i ( ( 𝑦𝐵𝑧𝐵 ) → ( ( 𝑦 𝐽 𝑧 ) × { ( 1𝑋 ) } ) = ( 𝑓 ∈ ( 𝑦 𝐽 𝑧 ) ↦ ( 1𝑋 ) ) )
14 13 mpoeq3ia ( 𝑦𝐵 , 𝑧𝐵 ↦ ( ( 𝑦 𝐽 𝑧 ) × { ( 1𝑋 ) } ) ) = ( 𝑦𝐵 , 𝑧𝐵 ↦ ( 𝑓 ∈ ( 𝑦 𝐽 𝑧 ) ↦ ( 1𝑋 ) ) )
15 11 14 opeq12i ⟨ ( 𝐵 × { 𝑋 } ) , ( 𝑦𝐵 , 𝑧𝐵 ↦ ( ( 𝑦 𝐽 𝑧 ) × { ( 1𝑋 ) } ) ) ⟩ = ⟨ ( 𝑦𝐵𝑋 ) , ( 𝑦𝐵 , 𝑧𝐵 ↦ ( 𝑓 ∈ ( 𝑦 𝐽 𝑧 ) ↦ ( 1𝑋 ) ) ) ⟩
16 10 15 eqtr4di ( 𝜑𝐾 = ⟨ ( 𝐵 × { 𝑋 } ) , ( 𝑦𝐵 , 𝑧𝐵 ↦ ( ( 𝑦 𝐽 𝑧 ) × { ( 1𝑋 ) } ) ) ⟩ )