Metamath Proof Explorer


Theorem qtopss

Description: A surjective continuous function from J to K induces a topology J qTop F on the base set of K . This topology is in general finer than K . Together with qtopid , this implies that J qTop F is the finest topology making F continuous, i.e. the final topology with respect to the family { F } . (Contributed by Mario Carneiro, 24-Mar-2015)

Ref Expression
Assertion qtopss FJCnKKTopOnYranF=YKJqTopF

Proof

Step Hyp Ref Expression
1 toponss KTopOnYxKxY
2 1 3ad2antl2 FJCnKKTopOnYranF=YxKxY
3 cnima FJCnKxKF-1xJ
4 3 3ad2antl1 FJCnKKTopOnYranF=YxKF-1xJ
5 simpl1 FJCnKKTopOnYranF=YxKFJCnK
6 cntop1 FJCnKJTop
7 5 6 syl FJCnKKTopOnYranF=YxKJTop
8 toptopon2 JTopJTopOnJ
9 7 8 sylib FJCnKKTopOnYranF=YxKJTopOnJ
10 simpl2 FJCnKKTopOnYranF=YxKKTopOnY
11 cnf2 JTopOnJKTopOnYFJCnKF:JY
12 9 10 5 11 syl3anc FJCnKKTopOnYranF=YxKF:JY
13 12 ffnd FJCnKKTopOnYranF=YxKFFnJ
14 simpl3 FJCnKKTopOnYranF=YxKranF=Y
15 df-fo F:JontoYFFnJranF=Y
16 13 14 15 sylanbrc FJCnKKTopOnYranF=YxKF:JontoY
17 elqtop3 JTopOnJF:JontoYxJqTopFxYF-1xJ
18 9 16 17 syl2anc FJCnKKTopOnYranF=YxKxJqTopFxYF-1xJ
19 2 4 18 mpbir2and FJCnKKTopOnYranF=YxKxJqTopF
20 19 ex FJCnKKTopOnYranF=YxKxJqTopF
21 20 ssrdv FJCnKKTopOnYranF=YKJqTopF