Does a surjective linear map commute with the interior of a convex body?

Question

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.

Answer 1

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.

Answer 2

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.