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Let $C \subset \mathbb{R}^{n}$ be a closed convex cone. If one wants to know whether the linear map $T:\mathbb{R}^{n} \to\mathbb{R}^m$ sends the closed set $C$ to another closed one, $T(C)$, it is needed to prove that $\text{ker } T + C$ is closed.

My concern turns into to know whether there exist good examples of cones, other than polyhedral cones, that are mapped to closed sets by any linear map. Hence, the question is reduced to:

There exist a closed convex cone $C$, different from any polyhedral cone, such that, for every vector subspace $V$, $V+C$ is closed?

The article On the closedness of the linear image of a closed convex cone almost answers my question. If one can find an enlightening example of cone $C$ satisfying the condition (SUM-WE) for every subspace, my example is constructed.

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    $\begingroup$ Since the linked article is behind a paywall it might be useful to formulate explicitely the SUM-WE condition. $\endgroup$ May 25 at 7:26
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    $\begingroup$ A copy of this article is available at the homepage of the author: gaborpataki.web.unc.edu/wp-content/uploads/sites/14119/2018/07/… $\endgroup$
    – gerw
    May 25 at 9:48
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    $\begingroup$ The fact that a closed convex cone is polyhedral iff all its projections are closed (which is essentially your question) was proved in 1957 in H.Mirkil, "New characterizations of polyhedral cones". See also the 1959 paper by V.Klee, "Some characterizations of convex polyhedra". $\endgroup$ May 25 at 13:53
  • $\begingroup$ @GuillaumeAubrun, there is a question by a student who heard one of your talks, something about a cone in $\mathbb R^3$ with a dense set of planar cross sections. math.stackexchange.com/questions/4469943/… Is there a standard reference for this sort of thing? $\endgroup$
    – Will Jagy
    Jun 10 at 21:40
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    $\begingroup$ I don't know of reference but had wrote up a note about this math.univ-lyon1.fr/homes-www/aubrun/recherche/… $\endgroup$ Jun 11 at 5:28

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The radial cone of $C$ is defined via $$ \mathcal R_C(x) := \bigcup_{\lambda > 0} \lambda ( C - x)$$ for all $x \in C$ and we can show $$ \mathcal R_C(x) = C + \operatorname{span}(x), $$ since $C$ is a cone. By assumption $\mathcal R_C(x)$ is closed for all $x \in C$, since $\operatorname{span}(x)$ is a subspace. But now, Proposition 2 of Duality of linear conic problems by Shapiro and Nemirovski implies that $C$ is a polyhedral cone.

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  • $\begingroup$ Holly mother of god! In the foundations of my pure being, and in my full essence, this is a completely unexpected result. God! That was unexpected! I was trully not hoping to find a "new" characterization of polyedral cones. $\endgroup$ May 25 at 12:46

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