Closed cone in the euclidean space $\mathbb{R}^n$ Let $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ be a linear map and let $\mathcal{C}$ be a closed cone in $\mathbb{R}^n.$ Prove that $T(\mathcal{C})$ is a closed cone in $\mathbb{R}^m$ provided $\ker(T)\cap \mathcal{C}=\{0\}$.
I have no problem to justify that $T(\mathcal{C})$ is a cone in $\mathbb{R}^m$. The issue is to verify that $T(\mathcal{C})$ is closed in $\mathbb{R}^m$. My idea is trying to use a  sequence characterization for closed sets, by which I mean that for any given sequence in $T(\mathcal{C})$ that converges to some $x\in\mathbb{R}^m$, then $x\in T(\mathcal{C}).$ But it was not successful, since I do not know how to apply the given hypothesis $\ker(T)\cap \mathcal{C}=\{0\}$. Does anyone have a useful option/thought or recommendation on this problem? Or if you have a better idea, I would be happy to listen to it.

Here is an example that shows what happens if the condition is violated: https://math.stackexchange.com/a/1457777/27978
Note that if $\ker T \cap C = \{ 0 \}$, then $\ker T \cap \operatorname{sp}C = \{ 0 \}$.
Define $A: {\cal R} T \to \operatorname{sp}C$ by $AT x = x$. 
Now suppose $t_k =Tc_k$, with $c_k \in C$ and $t_k \to t$. Since $At_k = c_k$ and $A t_k \to At$, we see that
$c_k \to At$, and since $C$ is closed, $At \in C$ and so $t \in TC$.
