$X$ compact convex subset of $\mathbb{R^n}$. $f:X\to X,f(Fr(X))\subset Fr(X)$. $f|_{Fr(X)}$ is homotopic with identity on $Fr(X)$. Show that $f(X)=X$ Let $X$ be compact convex subset of $\mathbb{R}^n$ with non-empty interior. $f:X\to X$ is map such that $f(Fr(X))\subset Fr(X)$. Moreover $f|_{Fr(X)}$ is homotopic in $Fr(X)$ with identity on $Fr(X)$. Show that $f(X)=X$.
I would like to use the fact that if $g:[0,1]^n\to[0,1]^n$ and $g(F_k)\subset F_k$ where $F_k$ is face of $[0,1]^n$, then $g([0,1]^n)=[0,1]^n$.
Assuming that $f(Fr(X))=Fr(X)$ this should be straightforward because $X$ is homeomorphic with $[0,1]^n$. However I do not know how to show that this assumption holds.
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Assuming that $f(Fr(X))\neq Fr(X)$, we get with other assumptions that $Fr(X)$ is homotopic with it's proper subset. It does not look right.
 A: All closed, bounded, convex subsets with non-empty interior are homeomorphic. One way is to show that each set is homeomorphic to the unit ball.
Suppose $X$ is closed, bounded and convex, with non-emtpy interior. Without loss of generality (by applying translation and scaling as necessary; all bicontinuous operations), we may assume that the open unit ball is contained in $X$.
Consider the gauge function of $X$:
$$\gamma(x) = \inf \lbrace \lambda \ge 0 : x \in \lambda X \rbrace.$$
This function is continuous, positive-definite, and takes values between $0$ and $1$ when mapping points inside $X$. In fact, $\gamma(x) = 1$ if and only if $x \in \operatorname{Fr}(X)$. Let $B$ be the closed unit ball, and define,
$$f : X \to B : x \mapsto \begin{cases}\frac{\gamma(x)}{\|x\|}x & \text{if } x \neq 0 \\ 0 & \text{if } x = 0\end{cases}.$$
It takes a little work, but one can show $f$ is a homeomorphism. I don't have a proof to hand, so I'll leave that to you to figure out the details.
A: You know that $X$ is homeomorphic to $[0,1]^n \approx D^n$. Let $h : X \to D^n$ be a homeomorphism. By invariance of domain we see that $h(int(X) = int(D^n)$, hence also $h(Fr(X)) = Fr(D^n) = S^{n-1}$. Let $f' = F\mid_{Fr(X)} : Fr(X) \to Fr(X)$ and $h' = h_{S^{n-1}} : S^{n-1} \to S^{n-1}$. Then $\phi = h' f' h^{-1}$ is homotopic to the identity.
If $f'$ would not be surjective, then we can pick $x \in Fr(X)$ with $x \notin f'(Fr(X))$. But then $\phi(S^{n-1}) \subset S^{n-1} \setminus \{ h'(x) \} \approx \mathbb R^{n-1}$. Hence $\phi$ is null-homotopic. This contradicts the fact that $\phi$ is homotopic to the identity.
