Show: $\ker(b) \cong \ker(a)/U$.

Question

I am working on the following task: Let $V,W$ be $K$-vector space, $a:V\to W$ is a homomorphism and $U \subset V$ a subspace, $U \subset \ker (a)$. Let $b: V/U\to W$, $v+U \mapsto a(v)$ be a homomorphism. Show: $\ker(b) \cong \ker(a)/U$. My idea: $pi: V \to V/U$, $v \mapsto v+U$.

Thanks for your help!

Answer 1

Hint : Show that the homomorphism $$\pi_{|\mathrm{Ker}(a)} : \mathrm{Ker}(a) \rightarrow \mathrm{Ker}(b)$$

is well-defined, surjective, and that its kernel is $U$. It will then induce an isomorphism $$\tilde{\pi}_{|\mathrm{Ker}(a)} : \mathrm{Ker}(a)/U \rightarrow \mathrm{Ker}(b)$$