Metamath Proof Explorer


Theorem fulli

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

Ref Expression
Hypotheses isfull.b 𝐵 = ( Base ‘ 𝐶 )
isfull.j 𝐽 = ( Hom ‘ 𝐷 )
isfull.h 𝐻 = ( Hom ‘ 𝐶 )
fullfo.f ( 𝜑𝐹 ( 𝐶 Full 𝐷 ) 𝐺 )
fullfo.x ( 𝜑𝑋𝐵 )
fullfo.y ( 𝜑𝑌𝐵 )
fulli.r ( 𝜑𝑅 ∈ ( ( 𝐹𝑋 ) 𝐽 ( 𝐹𝑌 ) ) )
Assertion fulli ( 𝜑 → ∃ 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) 𝑅 = ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑓 ) )

Proof

Step Hyp Ref Expression
1 isfull.b 𝐵 = ( Base ‘ 𝐶 )
2 isfull.j 𝐽 = ( Hom ‘ 𝐷 )
3 isfull.h 𝐻 = ( Hom ‘ 𝐶 )
4 fullfo.f ( 𝜑𝐹 ( 𝐶 Full 𝐷 ) 𝐺 )
5 fullfo.x ( 𝜑𝑋𝐵 )
6 fullfo.y ( 𝜑𝑌𝐵 )
7 fulli.r ( 𝜑𝑅 ∈ ( ( 𝐹𝑋 ) 𝐽 ( 𝐹𝑌 ) ) )
8 1 2 3 4 5 6 fullfo ( 𝜑 → ( 𝑋 𝐺 𝑌 ) : ( 𝑋 𝐻 𝑌 ) –onto→ ( ( 𝐹𝑋 ) 𝐽 ( 𝐹𝑌 ) ) )
9 foelrn ( ( ( 𝑋 𝐺 𝑌 ) : ( 𝑋 𝐻 𝑌 ) –onto→ ( ( 𝐹𝑋 ) 𝐽 ( 𝐹𝑌 ) ) ∧ 𝑅 ∈ ( ( 𝐹𝑋 ) 𝐽 ( 𝐹𝑌 ) ) ) → ∃ 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) 𝑅 = ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑓 ) )
10 8 7 9 syl2anc ( 𝜑 → ∃ 𝑓 ∈ ( 𝑋 𝐻 𝑌 ) 𝑅 = ( ( 𝑋 𝐺 𝑌 ) ‘ 𝑓 ) )