We have the following theorem (call it the universal mapping property for quotient spaces): $V,U$ are vector spaces over field $F$, $W⊂V$ a subspace. Then for every linear transformation $t:V→U$ so that $W⊂ker(t)$, there exists a unique linear transformation $T:V/W→U$ so that $T∘p=t$, where $p:V→V/W$ maps elements of $V$ to their cosets, i.e. $p(v)=v+W$.
Does it imply that there is a linear isomorphism from the subspace of linear operators in $\mathcal{L}(V,U)$ whose kernel contains $W$, and the space of linear operators $\mathcal{L}(V/W,U)$?
I guess that the theorem grants an injection from the former to the latter, since, if two functions $t_1$ and $t_2$ belonging to the former give rise to the same element of the latter, i.e., $T_1=T_2$, then we have, from the universal property and the assumption $T_1=T_2$ that $t_1=T_1∘p=T_2∘p=t_2$, which means $t_1=t_2$.
How would you show that instead the inverse map that takes a linear map $\phi$ in $\mathcal{L}(V/W,U)$ and associates to it the map $\psi$ defined by $\psi(v)=\phi(v+W)$ is is the inverse of the previous map?
I guess that one has to show that it takes value in the subspace of linear maps in $\mathcal{L}(V,U)$ whose kernel contains $W$, and that it is injective. Could you help me clarify if this is true and spell the last details out?
Thanks for all the help.
Just show that the map $L(V/W,U)\to L(V,U)$ that sends $\phi\mapsto \phi\circ p$ is injective, and its image is precisely those elements of $L(V,U)$ whose kernel contains $W$.