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Jul 23, 2019 at 8:45 vote accept Pavel
Jul 11, 2014 at 9:09 comment added Pavel In addition, $\partial_i E(f) = E(\partial_i f)$. It follows, if $f: \partial\mathbb{R}^n_+ \rightarrow \partial\mathbb{R}^n_+$ restricts to or. pres. diffeos of $\partial_i \mathbb{R}^n_+$ for all $i$ then $\mathrm{d}E(f)(x)$ is inverible whenever $x$ lies in at least two $\partial_i\mathbb{R}^n_+$. Here, $E$ is applied componentwisely. The inverse function theorem for mfds with corners asserts $E(f)$ is a diffeo of nbhds of $x$ and $f(x)$ in $\mathbb{R}^n_+$. If $x$ lies in precisely one $\partial_i\mathbb{R}^n_+$ it is easy to write down a diffeo loc. extending $f$. Hence, locally it works.
Jul 11, 2014 at 8:45 comment added Pavel Thank you very much for the reference. In fact, if $f: \partial \mathbb{R}^n_+ \rightarrow \mathbb{R}$ is a function smooth on $\partial_i \mathbb{R}^n_+ =\{x\in R^n_+ : x^i = 0\}$ for every $i$ then one can use the smooth extension $E(f)=\sum_{k=1}^n (-1)^{k+1}\sum_{\substack{I \subset \{1,\ldots,n\}}} f \circ \pi_I$ where $\pi_I$ is the obvious linear projection to $\partial_I \mathbb{R}^n_+ = \{x\in \mathbb{R}^n_+ : x^I = 0\}$. I guess this is what arises when one extends the Lee's inductive argument.
Jul 10, 2014 at 13:44 comment added Roberto Frigerio I am currently dealing with a different (but similar) issue. Here is a fact that is stated in Lee's book "Introduction to smooth manifolds". Lemma 16.8 says that, if $f\colon \partial\Delta \to M$ is a continuous map whose restriction to every face of $\Delta$ is smooth, then $f$ is smooth when considered as a map of the whole boundary $\partial \Delta$ into $M$. This statement does not deal with diffeomorphisms, but still ensures a smooth local extension of the $f_i$'s to a neighbourhood of $\partial \Delta$.
Jul 8, 2014 at 15:26 comment added Pavel Thank you very much for your helpful answer. I will keep thinking about the extension when $k < n + 1$ then. Btw. I do not understand the issue with smoothness at the corners: For instance, if $f, g: [0,1) \rightarrow \mathbb{R}$ are smooth functions with $f(0) = g(0)$ then $h(x,y) = f(x) + g(y) - f(0)$ is a smooth function $h: [0,1)^2 \rightarrow \mathbb{R}$ extending $f$ and $g$ from the boundary, right ? I can write similarly looking extensions in higher dimensions as well.
Jul 8, 2014 at 14:16 history answered Ryan Budney CC BY-SA 3.0