Does a surjective linear map commute with the interior of a convex body? Let $T:\Bbb R^n \to \Bbb R^m $ be a surjective linear map.
Suppose $A\subset\Bbb R^n$ is convex, compact, $0\in \operatorname{int}(A)$, and centrally symmetric.
Is it true that $T(\operatorname{int}(A))=\operatorname{int}(T(A))$?.
Notes: 


*

*$\operatorname{int}(S)$ means interior of $S$ in the topology of $\mathbb{R}^n$ or $\mathbb{R}^m$.

*Convexity is important: e.g., consider the image of sphere $S^2\subset \mathbb{R}^3$ under projection

*$A$ having nonempty interior is also important: consider the image of a disk $D^2\subset \mathbb{R}^3$ under the same projection.

 A: Since $T$ is surjective, it is open. Therefore $T(\operatorname{int} A)$ is an open subset of $T(A)$, so $T(\operatorname{int} A) \subseteq \operatorname{int} T(A)$.
Pick an $a_0 \in \operatorname{int} A$, and choose an arbitrary $y \in \bigl(\operatorname{int} T(A)\bigr) \setminus \{T(a_0)\}$. Then there is an $\varepsilon > 0$ such that $z = T(a_0) + (1+\varepsilon)\bigl(y - T(a_0)\bigr) \in \operatorname{int} T(A)$. Choose $x \in A \cap T^{-1}(\{z\})$. Then the "tipless cone"
$$C = \{ tx + (1-t)a : t \in [0,1),\, a \in \operatorname{int} A\}$$
is open, and by convexity contained in $A$, hence $C \subseteq \operatorname{int} A$. We have
$$w = a_0 + \frac{1}{1+\varepsilon}(x - a_0) \in C$$
and
\begin{align}
T(w) &= T(a_0) + \frac{1}{1+\varepsilon}\bigl(T(x) - T(a_0)\bigr) \\
&= T(a_0) + \frac{1}{1+\varepsilon}\bigl(z - T(a_0)\bigr) \\
&= T(a_0) + \frac{1}{1+\varepsilon}(1+\varepsilon)\bigl(y - T(a_0)\bigr) \\
&= y,
\end{align}
showing $y \in T(\operatorname{int} A)$. Since clearly also $T(a_0) \in T(\operatorname{int} A)$, it follows that $\operatorname{int} T(A) \subseteq T(\operatorname{int} A)$.
We note that the proof only used that $T$ is an open linear map, and $A$ convex with nonempty interior, so the result also holds in an infinite-dimensional setting.
A: The answer is 'yes'.  The proof is based on the lemma:
Lemma. Suppose $A$ is compact and convex in $\mathbb{R}^n$,
and $0$ is an interior point of $A$.
Let $R$ be a ray based at $0$.
Then $R$ intersects $\partial A$ in exactly one point $\bar{x}$.
All the points in the half-open segment 
$[0,\bar{x})$ are interior points of $A$.
Proof. See for example: the book Topology and Geometry by Bredon, p. 56.
And for your question:
It is easy to show that $T(A)$ is convex and compact.
The easy part:
Since $T$ is surjective, by the open mapping theorem,
$T( \mathrm{int}(A) )$ is open, and it is certainly a subset of $T(A)$, therefore 
$T( \mathrm{int}(A) ) \subseteq \mathrm{int}(T(A))$.
The hard part:
Conversely, suppose $y \in \mathrm{int}(T(A))$; we must show $y=T(a)$ for some $a \in \mathrm{int}(A)$.
By the open mapping theorem, $T(0)$ is an interior point of $T(A)$.
If $T(0)=y$ we are done.
Otherwise let $R_Y$ be the ray based at $T(0)$ through $y$.
By the Lemma, $R_Y$ intersects $\partial T(A)$ in a unique point $\bar{y}$.
Since $y$ is an interior point and $\bar{y}$ is a boundary point, $y \ne \bar{y}$.
Let $y_1$ be the midpoint of segment $[y,\bar{y})$.
By the Lemma, $y_1 \in \mathrm{int}(T(A)) \subseteq T(A)$.
Pick $a_1 \in A$ so $T(a_1)=y_1$; clearly $a_1 \ne 0$.
Consider the ray $R_X$ based at $0$ and passing through $a_1$.
$a_1$ is either an interior point or a boundary point of $A$,
but in either case, by the Lemma,
all the points in $[0,a_1)$ are interior points.
$T$ maps the segment $[0,a_1)$ to $[T(0),y_1)$.
Since the given $y \in [T(0),y_1)$,
one of those interior points in $[0,a_1)$ maps to $y$.
We are done.
Note that there is no assumption about symmetry.
