Quotient spaces and Subspaces
I need some help validating and completing my proof.
Let $K$ be a field and $U,T,V$ be Vector spaces over $K$ with the property $T\subseteq U \subseteq V $. Show that
- $U/T$ is a subspace of $V/T$.
- $(V/T)/ (U/T) \cong V/U$
I thought the first part was rather trivial, considering $T,U,V$ are all vector spaces I assumed that $U/T$ is nonempty and continued by using the definition of a quotient space:
$ \forall x,y \in U/T$ it is true that $[x]=x+T$ and $[y]=y+T $. Therefore $ x+y+T=[x+y]\in U/T$ . Since $U$ is a subspace of $V$ it follows that $x,y \in V$ and $[x+y]\in V/T $.
Proceeding in a similar way for scalar multiplication:
Let $a \in K $ then $ \forall x\in U/T $ then by definition of quotient space $a[x]=[ax]\in U/T$. Since $U \subseteq V$ it follows again that $a[x]=[ax] \in V/T $.
I believe this makes sense but I'm not entirely sure if it covers everything. Does it even make sense to assume $U/T$ isn't empty? By the second part I'm honestly a little clueless. I know that two vector spaces are isomorphic if there exists a bijective linear map between them but I don't know how that could be useful here.
Regarding the first part, your proof is perfectly fine. You can shorten your proof to the following. First, note that $U/T$ is a vector space by the definition of the quotient space. On the other hand, $U/T$ is also a subset of $V/T$ over which addition and scalar multiplication have the same meaning. Because $U/T$ is a subset of $V/T$ that is itself a vector space, it is by definition a subspace of $V/T$.
Regarding the second part, here is one such bijective linear map. Define $\phi:(V/T)/(U/T) \to V/U$ by $$ \phi((x + T) + U/T)) = x + U. $$ I will leave it to you to argue that this map is well defined (i.e. gives the same result regardless of which element $x$ is used to represent $x + T$), linear, and bijective.
Alternatively, you may have encountered the fact that for any linear transformation $f:V \to W$, we have $$ \operatorname{im}(f) \cong V/\ker(f). $$ With this in mind, you could also consider the map $f:V/T \to V/U$ given by $$ f(x + T) = x + U. $$ verify that this map is well defined, that it is surjective, and that its kernel is $U/T$.