The following is a theorem that I'm trying to prove:
Proof Idea:
So, I'm kind of stuck on this problem. Like, I don't have a proof but I have an idea for how it could go. I'm just having a hard time figuring out the last few steps.
Proving uniqueness really isn't the issue. In fact, in proving uniqueness, I got the idea that I could just define $\phi$ to be:
$$\phi = f \circ h^{-1}$$
where $h^{-1}:V/U \to V$ is a linear map that is the inverse of $h:V \to V/U$. In order for that inverse to exist, the idea I had is that $h:V \to V/U$ has to be a bijection.
Therefore, all I really have to do is to find a bijection between those two vector spaces. This didn't seem to be that hard, because I have a linear map that is an epimorphism between the two spaces given as follows:
$$\pi: V \to V/U$$
$$v \to v+U$$
It's already linear and surjective, so all I need to do is to prove injectivity. But I'm having a hard time doing that. I know that it has to do with the fact that $U \subset Ker(f)$ but I can't seem to figure out the way to use that fact in the proof.
So, could someone give me a hint on how I could continue? If my entire approach is wrong, how do I fix it?
I'd also appreciate it if you could write the proof for this as well. I'll look at your hint first and then try to construct my own proof. If I succeed in doing that, I'll have a look at your proof as a reference. Thanks in advance!
Edit:
Best Answer
Let $V$ be a vectorspace, $U$ a linear subspace of $V$. I will use your notation: We define $\varphi \colon V/U \rightarrow V$ via $\varphi(v+U) = f(v)$ for all $v \in V$. Then it is clear that $f \circ \pi = \varphi$. You now need to check that this is well-defined, that is, if $v +U = w + u$ then $f(u) = f(w)$. If $v+U = w +U$ , then $v -w \in U \subseteq \ker(f) \Rightarrow f(v) = f(v) - f(w) + f(w) = f(v-w) +f(w) = f(w)$ by linearity of $f$ and as $v-w \in \ker(f)$. I leave it to you to check that $\varphi$ is linear.