Subspace proof homomorphism Hom(V,W)

Question

Let $V$ and $W$ $K$-Vector space and $U$ a subspace of $V$. Prove for $H_U:= \{f ∈ \mathrm{Hom}(V,W); U ⊆ \ker f\}$

that:

$\mathrm{Hom}(V, W )/H_U \cong \mathrm{Hom}(U, W )$

Can someone explain it to me with a proof?

Answer 1

Define : $$\begin{matrix}\Phi&:&\text{Hom}(V,W)/H_U&\to&\text{Hom}(U,W)\\&&[f]&\mapsto&f|_U\end{matrix}.$$

$\Phi$ is well-defined, for if $[f]=[g]$, then $f-g\in H_U$, i.e. $(f-g)|_U=0$, i.e. $f|_U=g|_U$.

Now, $\ker(\Phi)=0$ : if $f|_U=0$, that means that $f\in H_U$, i.e. $[f]=0$. Finally, $\Phi$ is surjective, for if $g:U\to W$, then any linear continuation $\tilde{g}:V\to W$ will yield : $\Phi([\tilde{g}])=\tilde{g}|_U=g$.