Metamath Proof Explorer


Theorem fundcmpsurinj

Description: Every function F : A --> B can be decomposed into a surjective and an injective function. (Contributed by AV, 13-Mar-2024)

Ref Expression
Assertion fundcmpsurinj F:ABAVghpg:Aontoph:p1-1BF=hg

Proof

Step Hyp Ref Expression
1 abrexexg AVz|xAz=F-1FxV
2 1 adantl F:ABAVz|xAz=F-1FxV
3 fveq2 x=yFx=Fy
4 3 sneqd x=yFx=Fy
5 4 imaeq2d x=yF-1Fx=F-1Fy
6 5 eqeq2d x=yz=F-1Fxz=F-1Fy
7 6 cbvrexvw xAz=F-1FxyAz=F-1Fy
8 7 abbii z|xAz=F-1Fx=z|yAz=F-1Fy
9 8 fundcmpsurinjpreimafv F:ABAVghg:Aontoz|xAz=F-1Fxh:z|xAz=F-1Fx1-1BF=hg
10 foeq3 p=z|xAz=F-1Fxg:Aontopg:Aontoz|xAz=F-1Fx
11 f1eq2 p=z|xAz=F-1Fxh:p1-1Bh:z|xAz=F-1Fx1-1B
12 10 11 3anbi12d p=z|xAz=F-1Fxg:Aontoph:p1-1BF=hgg:Aontoz|xAz=F-1Fxh:z|xAz=F-1Fx1-1BF=hg
13 12 2exbidv p=z|xAz=F-1Fxghg:Aontoph:p1-1BF=hgghg:Aontoz|xAz=F-1Fxh:z|xAz=F-1Fx1-1BF=hg
14 2 9 13 spcedv F:ABAVpghg:Aontoph:p1-1BF=hg
15 exrot3 pghg:Aontoph:p1-1BF=hgghpg:Aontoph:p1-1BF=hg
16 14 15 sylib F:ABAVghpg:Aontoph:p1-1BF=hg