Question: let $V$, $V'$ be vector spaces over field $K$ and $W$ be subspace of $V$ then show that $\Gamma = \{T\in L(V,V')\vert\forall w\in W: T(w) = 0\}$ is subspace of $L(V,V')$. Further show that this subspace is isomorphic to $L(V/W,V')$ and if $V$, $V'$ are finite dimensional then $\dim\Gamma = (\dim V - \dim W)\dim V'$.
Note: $L(V,V')$ is vector space of all linear transformations from $V$ to $V'$.
My attempt: I had finished the first part. I had shown that $Γ$ is subspace of $L(V,V')$. Further I can see, $W$ is contained in $\ker T$ for all $T\in L(V,V')$. I know, I need to use isomorphism theorems. But I didn't able to go further. Please help me.
Hint: check the (obviously? linear) function $$f:\Gamma\longrightarrow L(V/W,V′)$$ $$T\longmapsto f(T)$$ $$f(T):V/W\longrightarrow V′$$ $$v + W\longmapsto T(v).$$ Why is well-defined? Why is isomorphism?...
Edit: about the well-definedness of $f(T)$: if $T\in \Gamma$, by definition, $$\forall w\in W: T(w) = 0.$$ This means that for all $v\in V, w\in W$: $T(v + w) = T(v)...$
Edit: about the injectivity of $F$ ($[] =$ class of equivalence):
$$f(T) = 0\iff\forall [v]\in V/W:\ T(v) = f(T) = 0\iff T = 0.$$ (why the last $\iff$?)
About the surjectivity: write $V$ as direct sum of $W$ with another subspace (call it $E$). Now taking $S\in L(V/W,V')$, we must define a $\bar S\in\Gamma$ with $f(\bar S) = S$. This means that for $v\in E:\ \bar S(v) =\cdots$