Quotient representation, in linear algebra and in topology A complex vector space $\,{\mathbb{V}}\,$ can be split into classes so that two vectors belong to the same class if their difference lies in a subspace $\,{\mathbb{V}}_B\subset \mathbb V\,$. By choosing a representative $\,v\,$ in a class, we can present the class as $\,v+{\mathbb{V}}_B\,$, $\,v\in{\mathbb{V}}\,$. These classes constitute a complex vector space named factor space and denoted with ${\mathbb{V}}/{\mathbb{V}}_B$.
If a representation $A(G)$ of a group $G$ is acting in $\,{\mathbb{V}}\,$, and its subrepresentation $B(G)$ is acting in $\,{\mathbb{V}}_B\,$, then in the factor space ${\mathbb{V}}/{\mathbb{V}}_B$ is acting a homomorphism
$$
 A/B\,:\;\;G\,\longrightarrow\,GL({\mathbb{V}}/{\mathbb{V}}_B)
 $$
$$
 (A/B)(g)\,(v\,+\,{\mathbb{V}}_B)\;\equiv\;A(g)\,v\;+\;{\mathbb{V}}_B
 $$
called quotient representation. Owing to the invariance of ${\mathbb{V}}_B$ under $A(G)$ or, equivalently, under $B(G)$, this definition is invariant under the choice of the representative $v$.
Now, suppose that $\,{\mathbb{V}}\,$ is a topological vector space (TVS) equipped with a topology $\cal T$ and written down as $\,({\mathbb{V}},\,\cal T)\,$.
In this case, ${\mathbb{V}}_B$ also will be a TVS, in the induced-topology sense.
Also, suppose that ${\mathbb{V}}_B$ is closed in $\,({\mathbb{V}},\,\cal T)\,$. This makes $B(G)$ a subrepresentation not only in the algebraic but also in the topological sense.
QUESTION
Will the factor space $A/B$ be a topological space?
MORE SPECIFICALLY
Is it pointless to ask if ${\mathbb{V}}/{\mathbb{V}}_B$ is topological in the sense of $\cal T\,$?
$\,$Will the topology $\cal T$ of ${\mathbb{V}}$ naturally induce a quotient topology ${\cal{T}}^{\prime}$ in the factor space ${\mathbb{V}}/{\mathbb{V}}_B$, so that ${\mathbb{V}}/{\mathbb{V}}_B$ could become a TVS space $({\mathbb{V}}/{\mathbb{V}}_B\,,\;{\cal{T}}^{\prime})\,$?
 A: One can always form a space with the so-called quotient topology.
In general, one can take a set $S$ and quotient by an equivalence relation $\sim \subseteq S^2$ to get a map and a set $\pi : S \to S / \sim$.
If $S$ has a topology, then $S / \sim$ can be given the quotient topology, where a set $U \subseteq S / \sim$ is open iff $\pi^{-1}(U)$ is open. This topology makes $S / \sim$ into a topological space and $\pi : S \to S / \sim$ into a continuous map.
If we're dealing with vector space quotients, then we know the quotient map $\pi : \mathbb{V} \to \mathbb{V} / \mathbb{V}_B$ is a linear map. We can then topologise $\mathbb{V} / \mathbb{V}_B$ with the quotient topology. One can then prove that $\mathbb{V} / \mathbb{V}_B$ is a topological vector space under this topology and the induced vector space structure.
In fact, $\mathbb{V} / \mathbb{V}_B$ has the following property:
Let $f : \mathbb{V} \to U$ be a continuous linear map between topological vector spaces such that $\mathbb{V}_B \subseteq \ker f$. Then $f$ extends to a unique map $f' : \mathbb{V} / \mathbb{V}_B \to U$ such that $f' \circ \pi = f$; furthermore, $f'$ is a continuous linear map.
This is just piecing together the properties of quotient sets (which state that $f'$ exists), quotient vector spaces (which state that $f'$ is linear), and quotient topological spaces (which state that $f'$ is continuous).
In particular, consider a representation $A : G \to Hom_{TopVec}(\mathbb{V}, \mathbb{V})$ which takes an element of the group $G$ and outputs a continuous linear map $\mathbb{V} \to \mathbb{V}$. In particular, let us assume that it restricts to a map $\mathbb{V}_B$ to $\mathbb{V}_B$.
Note that given $g \in G$, we see that the map $\pi \circ A_g$ has the property that $\mathbb{V}_B \subseteq \ker (\pi \circ A_g)$. Thus, this map extends to a unique continuous linear map $\mathbb{V}_B \to \mathbb{V}_B$. It's not hard to show that this results in a representation $A / B : G \to Hom_{TopVec}(\mathbb{V} / \mathbb{V}_B, \mathbb{V} / \mathbb{V}_B)$.
Note: if you require a topological vector space to be $T_1$ then you need $\mathbb{V}_B$ to be closed in order for the quotient space to be $T_1$.
I'm not sure what you mean when you ask whether $A / B$ itself is a topological space. I thought that $A / B$ was supposed to be a representation.
The last thing to note here is that we don't necessarily need the underlying field of our vector spaces to be $\mathbb{C}$. In fact, any topological field will do. If you'd prefer to work with a field that doesn't have a particularly special topology, you can topologise it with the discrete topology.
