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Fixed typo, as suggested by Tom Werner
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Robert Furber
Robert Furber
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The question in the title is not how I'd phrase your question. I'd say what you're asking is "Is the preimage of a face under an affine map a face?", and the answer to that is yes.

The argument is simple. Let $f : X \rightarrow Y$ be an affine map (in your case, $\pi : X \rightarrow \pi(X)$), and let $F$ be a face of $Y$. To show that $f^{-1}(F)$ is a face of $X$, let $x,x' \in X$ and $\alpha \in [0,1]$ such that the convex combination $\alpha x + (1 -\alpha)x' \in f^{-1}(F)$. Because $f$ is affine, $\alpha f(x) + (1 - \alpha)f(x') \in F$. We then use the fact that $F$ is a face to deduce that $f(x), f(x') \in F$, and therefore $x,x' \in F$$x,x' \in f^{-1}(F)$, proving $F$$f^{-1}(F)$ is a face. (In case of confusion, I will state that I consider the empty set to be a face.)

For completeness, I'll say that the answer to the question I get from a natural interpretation of your title is no (is the projection of a face a face). Take $X$ to be an equilateral triangle and take $\pi$ to be the projection onto its base $Y$. The apex of the triangle is an extreme point, and therefore a face of $X$, but it projects down to a point half-way between the two extreme points of the base $Y$, which is not a face of $Y$.

The question in the title is not how I'd phrase your question. I'd say what you're asking is "Is the preimage of a face under an affine map a face?", and the answer to that is yes.

The argument is simple. Let $f : X \rightarrow Y$ be an affine map (in your case, $\pi : X \rightarrow \pi(X)$), and let $F$ be a face of $Y$. To show that $f^{-1}(F)$ is a face of $X$, let $x,x' \in X$ and $\alpha \in [0,1]$ such that the convex combination $\alpha x + (1 -\alpha)x' \in f^{-1}(F)$. Because $f$ is affine, $\alpha f(x) + (1 - \alpha)f(x') \in F$. We then use the fact that $F$ is a face to deduce that $f(x), f(x') \in F$, and therefore $x,x' \in F$, proving $F$ is a face. (In case of confusion, I will state that I consider the empty set to be a face.)

For completeness, I'll say that the answer to the question I get from a natural interpretation of your title is no (is the projection of a face a face). Take $X$ to be an equilateral triangle and take $\pi$ to be the projection onto its base $Y$. The apex of the triangle is an extreme point, and therefore a face of $X$, but it projects down to a point half-way between the two extreme points of the base $Y$, which is not a face of $Y$.

The question in the title is not how I'd phrase your question. I'd say what you're asking is "Is the preimage of a face under an affine map a face?", and the answer to that is yes.

The argument is simple. Let $f : X \rightarrow Y$ be an affine map (in your case, $\pi : X \rightarrow \pi(X)$), and let $F$ be a face of $Y$. To show that $f^{-1}(F)$ is a face of $X$, let $x,x' \in X$ and $\alpha \in [0,1]$ such that the convex combination $\alpha x + (1 -\alpha)x' \in f^{-1}(F)$. Because $f$ is affine, $\alpha f(x) + (1 - \alpha)f(x') \in F$. We then use the fact that $F$ is a face to deduce that $f(x), f(x') \in F$, and therefore $x,x' \in f^{-1}(F)$, proving $f^{-1}(F)$ is a face. (In case of confusion, I will state that I consider the empty set to be a face.)

For completeness, I'll say that the answer to the question I get from a natural interpretation of your title is no (is the projection of a face a face). Take $X$ to be an equilateral triangle and take $\pi$ to be the projection onto its base $Y$. The apex of the triangle is an extreme point, and therefore a face of $X$, but it projects down to a point half-way between the two extreme points of the base $Y$, which is not a face of $Y$.

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Robert Furber
Robert Furber
  • 3.8k
  • 1
  • 23
  • 34

The question in the title is not how I'd phrase your question. I'd say what you're asking is "Is the preimage of a face under an affine map a face?", and the answer to that is yes.

The argument is simple. Let $f : X \rightarrow Y$ be an affine map (in your case, $\pi : X \rightarrow \pi(X)$), and let $F$ be a face of $Y$. To show that $f^{-1}(F)$ is a face of $X$, let $x,x' \in X$ and $\alpha \in [0,1]$ such that the convex combination $\alpha x + (1 -\alpha)x' \in f^{-1}(F)$. Because $f$ is affine, $\alpha f(x) + (1 - \alpha)f(x') \in F$. We then use the fact that $F$ is a face to deduce that $f(x), f(x') \in F$, and therefore $x,x' \in F$, proving $F$ is a face. (In case of confusion, I will state that I consider the empty set to be a face.)

For completeness, I'll say that the answer to the question I get from a natural interpretation of your title is no (is the projection of a face a face). Take $X$ to be an equilateral triangle and take $\pi$ to be the projection onto its base $Y$. The apex of the triangle is an extreme point, and therefore a face of $X$, but it projects down to a point half-way between the two extreme points of the base $Y$, which is not a face of $Y$.