Metamath Proof Explorer


Theorem indlcim

Description: An independent, spanning family extends to an isomorphism from a free module. (Contributed by Stefan O'Rear, 26-Feb-2015)

Ref Expression
Hypotheses indlcim.f F = R freeLMod I
indlcim.b B = Base F
indlcim.c C = Base T
indlcim.v · ˙ = T
indlcim.n N = LSpan T
indlcim.e E = x B T x · ˙ f A
indlcim.t φ T LMod
indlcim.i φ I X
indlcim.r φ R = Scalar T
indlcim.a φ A : I onto J
indlcim.l φ A LIndF T
indlcim.s φ N J = C
Assertion indlcim φ E F LMIso T

Proof

Step Hyp Ref Expression
1 indlcim.f F = R freeLMod I
2 indlcim.b B = Base F
3 indlcim.c C = Base T
4 indlcim.v · ˙ = T
5 indlcim.n N = LSpan T
6 indlcim.e E = x B T x · ˙ f A
7 indlcim.t φ T LMod
8 indlcim.i φ I X
9 indlcim.r φ R = Scalar T
10 indlcim.a φ A : I onto J
11 indlcim.l φ A LIndF T
12 indlcim.s φ N J = C
13 fofn A : I onto J A Fn I
14 10 13 syl φ A Fn I
15 3 lindff A LIndF T T LMod A : dom A C
16 11 7 15 syl2anc φ A : dom A C
17 16 frnd φ ran A C
18 df-f A : I C A Fn I ran A C
19 14 17 18 sylanbrc φ A : I C
20 1 2 3 4 6 7 8 9 19 frlmup1 φ E F LMHom T
21 1 2 3 4 6 7 8 9 19 islindf5 φ A LIndF T E : B 1-1 C
22 11 21 mpbid φ E : B 1-1 C
23 1 2 3 4 6 7 8 9 19 5 frlmup3 φ ran E = N ran A
24 forn A : I onto J ran A = J
25 10 24 syl φ ran A = J
26 25 fveq2d φ N ran A = N J
27 23 26 12 3eqtrd φ ran E = C
28 dff1o5 E : B 1-1 onto C E : B 1-1 C ran E = C
29 22 27 28 sylanbrc φ E : B 1-1 onto C
30 2 3 islmim E F LMIso T E F LMHom T E : B 1-1 onto C
31 20 29 30 sylanbrc φ E F LMIso T