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 : A B A V g h p g : A onto p h : p 1-1 B F = h g

Proof

Step Hyp Ref Expression
1 abrexexg A V z | x A z = F -1 F x V
2 1 adantl F : A B A V z | x A z = F -1 F x V
3 fveq2 x = y F x = F y
4 3 sneqd x = y F x = F y
5 4 imaeq2d x = y F -1 F x = F -1 F y
6 5 eqeq2d x = y z = F -1 F x z = F -1 F y
7 6 cbvrexvw x A z = F -1 F x y A z = F -1 F y
8 7 abbii z | x A z = F -1 F x = z | y A z = F -1 F y
9 8 fundcmpsurinjpreimafv F : A B A V g h g : A onto z | x A z = F -1 F x h : z | x A z = F -1 F x 1-1 B F = h g
10 foeq3 p = z | x A z = F -1 F x g : A onto p g : A onto z | x A z = F -1 F x
11 f1eq2 p = z | x A z = F -1 F x h : p 1-1 B h : z | x A z = F -1 F x 1-1 B
12 10 11 3anbi12d p = z | x A z = F -1 F x g : A onto p h : p 1-1 B F = h g g : A onto z | x A z = F -1 F x h : z | x A z = F -1 F x 1-1 B F = h g
13 12 2exbidv p = z | x A z = F -1 F x g h g : A onto p h : p 1-1 B F = h g g h g : A onto z | x A z = F -1 F x h : z | x A z = F -1 F x 1-1 B F = h g
14 2 9 13 spcedv F : A B A V p g h g : A onto p h : p 1-1 B F = h g
15 exrot3 p g h g : A onto p h : p 1-1 B F = h g g h p g : A onto p h : p 1-1 B F = h g
16 14 15 sylib F : A B A V g h p g : A onto p h : p 1-1 B F = h g