Inducing a Linear Transformation on the Quotient Space Consider $W\subseteq V$, a subspace over a field $\mathbb{F}$ and $T:V\rightarrow V$ a linear transformation with the stipulation that $T(W)\subseteq W$.  Then we have the induced linear transformation $\overline{T}:V/W \rightarrow V/W$ such that $\overline{T}=T(v)+W$.
I'm supposed to show that this induced transformation is well-defined, and that given $V$ finite and $T$ an isomorphism that $\overline{T}$ is an isomorphism.  I'm having a little trouble with this part.  Namely, I want to show that the $ker(\overline{T})=W$, and I'm to the point where I realize that this means $T(v)\in W$.  How do I know there's not some random $v\in V\setminus W $ such that $ T(v)\in W$.
In particular, is all this true if $V$ is not finite dimensional?  I can't immediately think of a counterexample...
 A: Hint:  Given a linear isomorphism $T : V \rightarrow V'$ of finite-dimensional vector spaces, can you prove that the restricted linear transformation $T|_W : W \rightarrow T(W)$ is an isomorphism?
If $V' = V$ and $T(W) \subseteq W$ as in your problem, can you prove that $T(W) = W$?  In that case, you'll have $T^{-1}(W) = W$, which may prove useful in showing that $\ker \overline{T} = \{0\}$.
A: When you restrict to $W$, you still get a linear transformation -- this much is clear. Since $T$ is injective, its restriction to $W$ is still injective. Therefore, the dimension of the image $T(W)$ is equal to the dimension of $W$. Since $T(W)\subseteq W$, it readily follows that we have an equality. Since $T|_W$ is both injective and surjective, it is an isomorphism.
A few things change when you go to the infinite-dimensional case. First of all, it may be the case that $T(W)\subseteq W$, without equality. For example, this could happen if $W$ is a closed subset of $V$, but $T(W)$ isn't. Also, if topology is involved, a linear isomorphism may not be the kind of isomorphism you are looking for -- you may want bicontinuity as well.
