Metamath Proof Explorer


Theorem lmhmkerlss

Description: The kernel of a homomorphism is a submodule. (Contributed by Stefan O'Rear, 1-Jan-2015)

Ref Expression
Hypotheses lmhmkerlss.k ⊢ K = F -1 0 ˙
lmhmkerlss.z ⊢ 0 ˙ = 0 T
lmhmkerlss.u ⊢ U = LSubSp ⁡ S
Assertion lmhmkerlss ⊢ F ∈ S LMHom T → K ∈ U

Proof

Step Hyp Ref Expression
1 lmhmkerlss.k ⊢ K = F -1 0 ˙
2 lmhmkerlss.z ⊢ 0 ˙ = 0 T
3 lmhmkerlss.u ⊢ U = LSubSp ⁡ S
4 lmhmlmod2 ⊢ F ∈ S LMHom T → T ∈ LMod
5 eqid ⊢ LSubSp ⁡ T = LSubSp ⁡ T
6 2 5 lsssn0 ⊢ T ∈ LMod → 0 ˙ ∈ LSubSp ⁡ T
7 4 6 syl ⊢ F ∈ S LMHom T → 0 ˙ ∈ LSubSp ⁡ T
8 3 5 lmhmpreima ⊢ F ∈ S LMHom T ∧ 0 ˙ ∈ LSubSp ⁡ T → F -1 0 ˙ ∈ U
9 7 8 mpdan ⊢ F ∈ S LMHom T → F -1 0 ˙ ∈ U
10 1 9 eqeltrid ⊢ F ∈ S LMHom T → K ∈ U