The question is:
Let $T : \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a linear map. Let $C$ be a closed subset of $\mathbb{R}^n$ which is also a cone. Prove that if $\ker(T) \cap C = \{0\}$ then $T(C)$ is a closed cone in $\mathbb{R}^m$.
I can show that $T(C)$ is a cone but I can't figure out how to show it is closed.
This question was on my PhD qualifying exam I took recently and I'm going back over it to study.
If the map is surjective it is obvious because in that case
$T^\sim : \mathbb{R}^n/ker(T)\to \mathbb{R}^m$ is an omeomorphism (by open map theorem because $T$ is linear) and so $T^\sim(\pi(C)) $ is closed if and only if $ (T^\sim)^{-1}(T^\sim(\pi(C))=\pi(C)$ is closed in the quotient space that it is closed because $C$ is closed (infact $\pi^{-1}(\pi(C)=C $ if you have that $T$ satisfies the following condition:
If $T(x)\in T(C)$ then $x\in C$ )
So $T^\sim (\pi(C))$ is closed in $\mathbb{R}^m$ but you can observe that
$T^\sim(\pi(C))=T(C)$ because
if $y\in T^\sim (\pi(C))$ then there exist $\pi(x)\in \pi(C)$ such that
$y=T^\sim(\pi(x))$
You have that $\pi(x)\in \pi(C)$ so there exist $a\in C$ such that $\pi(x)=\pi(a)$ so $x-a\in ker(T)$ and you have that
$T(x)=T(a)$
Then $y=T(a)\in T(C)$
If $T(x)\in T(C)$ with $x\in C$ then it is clear that
$T(x)=T^\sim(\pi(x))$ so
$T^\sim(\pi(C))=T(C)$
