Metamath Proof Explorer


Theorem monsect

Description: If F is a monomorphism and G is a section of F , then G is an inverse of F and they are both isomorphisms. This is also stated as "a monomorphism which is also a split epimorphism is an isomorphism". (Contributed by Mario Carneiro, 3-Jan-2017)

Ref Expression
Hypotheses sectmon.b B=BaseC
sectmon.m M=MonoC
sectmon.s S=SectC
sectmon.c φCCat
sectmon.x φXB
sectmon.y φYB
monsect.n N=InvC
monsect.1 φFXMY
monsect.2 φGYSXF
Assertion monsect φFXNYG

Proof

Step Hyp Ref Expression
1 sectmon.b B=BaseC
2 sectmon.m M=MonoC
3 sectmon.s S=SectC
4 sectmon.c φCCat
5 sectmon.x φXB
6 sectmon.y φYB
7 monsect.n N=InvC
8 monsect.1 φFXMY
9 monsect.2 φGYSXF
10 eqid HomC=HomC
11 eqid compC=compC
12 eqid IdC=IdC
13 1 10 11 12 3 4 6 5 issect φGYSXFGYHomCXFXHomCYFYXcompCYG=IdCY
14 9 13 mpbid φGYHomCXFXHomCYFYXcompCYG=IdCY
15 14 simp3d φFYXcompCYG=IdCY
16 15 oveq1d φFYXcompCYGXYcompCYF=IdCYXYcompCYF
17 14 simp2d φFXHomCY
18 14 simp1d φGYHomCX
19 1 10 11 4 5 6 5 17 18 6 17 catass φFYXcompCYGXYcompCYF=FXXcompCYGXYcompCXF
20 1 10 12 4 5 11 6 17 catlid φIdCYXYcompCYF=F
21 1 10 12 4 5 11 6 17 catrid φFXXcompCYIdCX=F
22 20 21 eqtr4d φIdCYXYcompCYF=FXXcompCYIdCX
23 16 19 22 3eqtr3d φFXXcompCYGXYcompCXF=FXXcompCYIdCX
24 1 10 11 4 5 6 5 17 18 catcocl φGXYcompCXFXHomCX
25 1 10 12 4 5 catidcl φIdCXXHomCX
26 1 10 11 2 4 5 6 5 8 24 25 moni φFXXcompCYGXYcompCXF=FXXcompCYIdCXGXYcompCXF=IdCX
27 23 26 mpbid φGXYcompCXF=IdCX
28 1 10 11 12 3 4 5 6 17 18 issect2 φFXSYGGXYcompCXF=IdCX
29 27 28 mpbird φFXSYG
30 1 7 4 5 6 3 isinv φFXNYGFXSYGGYSXF
31 29 9 30 mpbir2and φFXNYG