Vector Space: is there a general definition of annihilator beyond the dual space? The definitions I see for the annihilator of a subset S of a vector space V over a field F is the subset $S^0$ of the dual space $V^*$given by $S^0 = \{\varphi\in V^\star:\varphi(S)=0\}$, and the annihilator can be seen to be a subspace.
My question is whether this is ever extended to a more general situation ? The dual space corresponds to $Hom(V, F)$ the set of linear transformations from V to F. So if instead of $Hom(V, F)$ one considers $Hom(V, W)$, the set of linear transformations from V to a space W (over the same field) it seems there should be a corresponding subspace of $Hom(V, W)$ which maps the subset S of V to the zero vector in W.
I'd appreciate any feedback on whether this is in fact the case and the terminology that applies.
 A: You have
$$
S^0=\{\varphi\in V^*:\varphi(S)=\{0\}\}=
\{\varphi\in V^*:S\subseteq\ker\varphi\}=
\{\varphi\in V^*:\langle S\rangle\subseteq\ker\varphi\}
$$
so it's not really restrictive to suppose $S$ is a subspace to begin with.
So, if $U$ is a subspace of $V$, then $U^0$ is the set of linear forms $\varphi\colon V\to F$ that factor through the canonical projection $V\to V/U$. As a consequence, you can identify $U^0$ with $(V/U)^*$.
It's surely possible to extend this to the more general setting, of course with $W\ne0$: if $U$ is a subspace of $V$, then
$$
U^{0,W}=\{\varphi\in\operatorname{Hom}_F(V,W):\varphi(U)=\{0\}\}
$$
can be “canonically” identified with $\operatorname{Hom}_F(V/U,W)$, with exactly the same argument as before.
The only missing point in the “identification” is the easy fact that if we have a proper subspace $U$ of $V$ and $v\in V\setminus U$, then there exists a linear map $\varphi\colon V\to W$ (in particular it can be $W=F$) such that $\varphi(U)=0$ and $\varphi(v)\ne0$.
Just consider a basis for $U$, add to it $v$ to get a linearly independent set and extend it to a basis of $V$; then defining the required linear map is easy.
Note that finite dimensionality is not needed in this context.

I assume you know about the quotient space, which is an indispensable tool when talking about duality. If $U$ is a subspace of $V$, we can consider the canonical projection $\pi_U\colon V\to V/U$. A linear map $f\colon V\to W$ factors through the canonical projection $\pi_U$ if there exists a linear map $g\colon V/U\to W$ such that
$$
f=g\circ \pi_U
$$
It is clear that, in this case, $\ker f\supseteq U$ (because $\ker\pi_U=U$). Conversely, if $U\subseteq \ker f$, then we can define $g\colon V/U\to W$ with $g(\pi_U(v))=f(v)$ and this is a good definition of a (linear) mapping, because if $\pi_U(v)=\pi_U(v')$, then $v-v'\in U$ and so $f(v)-f(v')=f(v-v')=0$. 
Note that $g$ is uniquely determined by $f$.
This factorization provides a (linear) mapping $U^0\to (V/U)^*$, because if $\varphi\in U^0$ we know there exists $\psi\colon V/U\to F$ with $\psi\circ\pi_U=\varphi$. This mapping is obviously surjective, but also injective: if $\varphi\in U^0$, $\varphi=\psi\circ\pi_U$ and $\psi=0$, then also $\varphi=0$.
