Example of homeomorphism that lifts to real blow up but not C^1? Given smooth manifold $M$, let $Bl_\Delta(M\times M)$ be the (say oriented; you can ask this question for the unoriented case too) real blow up of $M\times M$ along the diagonal and let $\pi:Bl_\Delta(M\times M)\to M\times M$ denote the blow down map.
Let $G(M)$ be the space of homeomorphisms $f:M\to M$ satisfying: there exists a homeomorphism $\tilde{f}:Bl_\Delta(M\times M)\to Bl_\Delta(M\times M)$ lifting $(f,f):M\times M\to M\times M$, i.e., $\pi\circ\tilde{f}=(f,f)\circ\pi$.
Clearly $\{C^1\text{-homeomorphisms of }M\}\subset G(M)\subset \text{Homeo}(M)$ and it is easy to give examples showing $G(M)\neq\text{Homeo}M$. My question is: is $G(M)$ the same as the space of $C^1$-homeomorphisms of $M$? I have trouble coming up with counterexamples. My only counterexample is when $M$ is 1-dimensional, in which case it is easy. But are there counterexamples if we assume $\dim(M)\ge2$?
Locally, a (continuous, injective) map $f:\mathbb{R}^n\to\mathbb{R}^n$ satisfies the lifting condition above implies that: there exists a homeomorphism $\phi\in\text{Homeo}(S^{n-1})$, such that for any sequance of pairs of points $\{(x_i,y_i)\in\mathbb{R}^n\times\mathbb{R}^n\}_{i=1}^\infty$, $$\lim_{i\to\infty}x_i,y_i=0,\ \lim_{i\to\infty}\frac{y_i-x_i}{|y_i-x_i|}=v\in S^{n-1}\implies\lim_{i\to\infty}\frac{f(y_i)-f(x_i)}{|f(y_i)-f(x_i)|}=\phi(v).$$
Are there such $f$'s that are not $C^1$ at 0?
 A: Ok, here's an example: $$F:B=B^n_{1/(2e)}\subset\mathbb{R}^n\to\mathbb{R}^n,F(x)=-2\log(|x|)\cdot x,$$ where $B^n_{\epsilon}$ is the $n$-ball with radius $\epsilon$, and $F$ maps $B^n_{1/(2e)}$ homeomorphically onto its image.
(We can extend $F$ smoothly to the whole $\mathbb{R}^n$ to make a homeomorphism.) Below is a detailed write up showing $F$ is as required.

*

*$F$ is not differentiable at 0: assume it is, then the differential must be a scaling, say by $k\in\mathbb{R}$. Then $\lim_{x\to0}\frac{F(x)-kx}{|x|}=\lim_{x\to0}\frac{-2\log|x|-k}{|x|}\cdot x=0$, but $\lim_{x\to0}\frac{\log|x|}{|x|}=\infty$, a contradiction.

Since $F$ is $C^1$ away from 0, $(F,F):B^n\times B^n\to\mathbb{R}^n\times\mathbb{R}^n$ lifts to $Bl_\Delta(B\times B)-\pi^{-1}(0,0)$. Define
$$\widetilde{F}:Bl_\Delta(B\times B)\to Bl_\Delta(\mathbb{R}^n\times\mathbb{R}^n)$$
to be $(F,F)$ away from $\pi^{-1}(0,0)$ and identity on $\pi^{-1}(0,0)$. To show it is continuous, we show for any sequence of points $\{p_i\in Bl_\Delta(B\times B)\}_{i=1}^\infty$ and $p\in\pi^{-1}(0,0)$, $p_i\to p\implies\widetilde{F}(p_i)\to\widetilde{F}(p)$. By writing $\{p_i\}$ as the union of subsequences $\{q^1_i\},\{q^2_i\},\{q^{3}_i\}$ where $q^1_i\in\pi^{-1}(B\times B-\Delta)$, $q^2_i\in\pi^{-1}(\Delta-(0,0))$, $q^{3}_i\in\pi^{-1}(0,0)$, it suffices to show for each $j=1,2,3$, $\lim_{i\to\infty}q^j_i=p\implies\lim_{i\to\infty}\widetilde{F}(q^j_i)=\widetilde{F}(p)$. This is clear for $j=3$. For $j=1,2$ it translates to the following two statements, respectively.
(1) For any $\{(x_i,y_i)\in B\times B\}_{i=1}^\infty$ such that $x_i\neq y_i$, $(x_i,y_i)\to(0,0)$, $$\lim_{i\to\infty}\frac{y_i-x_i}{|y_i-x_i|}=p\in S^{n-1}\implies\lim_{i\to\infty}\frac{F(y_i)-F(x_i)}{|F(y_i)-F(x_i)|}=p.$$
(2) For any $\{(x_i,v_i)\in (B-0)\times S^{n-1}\}_{i=1}^\infty$ such that $x_i\to 0$,$v_i\to p$,
$$\lim_{i\to\infty}(d_{x_i}F)(v_i)=p.$$
We check below that $F$ satisfies these two conditions.

*

*Denote $e_1=\frac{\partial}{\partial x_1}\in S^{n-1}$ the unit vector parallel to the $x_1$-axis. Since for any rotation $r$ of $\mathbb{R}^n$, $F\circ r=r\circ F$ and $r$ is linear, we have (identifying the tangent space of $\mathbb{R}^n$ at each point with $\mathbb{R}^n$) $d_xF\circ r=r\circ d_xF$ for any $x\in B$.
Therefore it would suffice to check the statement (1) for sequences of pairs of points $\{(x_i,y_i)\}_i$ such that $\frac{y_i-x_i}{|y_i-x_i|}=e_1$ (respectively, statement (2) for $\{(x_i,v_i)\}_i$ such that $v_i\equiv e_1$): if $\{(x_i,y_i)\}_i$ is such that $\lim_{i\to\infty}\frac{y_i-x_i}{|y_i-x_i|}=e_1$, then let $r_i$ be the rotation such that $r_i(y_i-x_i)$ is parallel to $e_1$; since
$$r_i(F(y_i)-F(x_i))=r_i(F(y_i))-r_i(F(x_i))=F(r_i(y_i))-F(r_i(x_i)),$$
if the direction of the last term converges to $e_1$, then so is the direction of $F(y_i)-F(x_i)$, since $||r_i||\to0$ where $||r_i||$ is the $L^\infty$-norm of $r_i:S^{n-1}\to S^{n-1}$. The case of (2) is similar.


*Abusing notation we also write $e_1$ for the tangert vector in the $e_1$ direction at any point $x\in\mathbb{R}^n$. For $x\neq0$ let $s(x)=\frac{|d_xF(e_1)-\langle d_xF(e_1),e_1\rangle e_1|}{\langle d_xF(e_1),e_1\rangle}$ be the "slope" of $(d_xF)(e_1)$. We show $\lim_{x\to0}s(x)=0$: direct computation shows
$$s(x)=\frac{2x_1\sqrt{x^2_2+\ldots+x^2_n}}{2x_1^2+|x|^2\log(|x|^2)};$$
so for $|x|$ small enough so that $\log|x|^2<-2$,
$$|s(x)|=\frac{2|x_1|\sqrt{x_2^2+\ldots+x_n^2}}{|x|^2(-\log|x|^2)-2x_1^2}<\frac{2|x_1|\sqrt{x_2^2+\ldots+x_n^2}}{|x|^2(-\log|x|^2-2)}<\frac{1}{-\log|x|^2-2}\to0$$
as $x\to0$, where the last inequality follows from Cauchy's inequality. This proves (2).


*Let $\{(x_i,y_i)\in\mathbb{R}^n\times\mathbb{R}^n\}_{i=1}^\infty$ be such that $\frac{y_i-x_i}{|y_i-x_i|}=e_1, \lim_{i\to\infty}x_i,y_i=0$. Denote by $F_1:\mathbb{R}^n\to\mathbb{R}$ the first factor of $F$ and $G:\mathbb{R}^n\to\mathbb{R}^{n-1}$ the other factors. We show below that $$\lim_{i\to\infty}\frac{|G(y_i)-G(x_i)|}{|F_1(y_i)-F_1(x_i)|}=0$$
which would conclude the proof. Given $\epsilon>0$, let $\delta>0$ be such that $|s(x)|<\epsilon$ for all $x$ with $|x|<\delta$. We check $|G(y_i)-G(x_i)|<\epsilon\,|F_1(y_i)-F_1(x_i)|$ whenever $|x_i|,|y_i|<\delta$. Suppose the first coordinate of $x_i$ is smaller than that of $y_i$. Let $\gamma:[a,b]\to\mathbb{R}^n,\gamma(a)=x_i,\gamma(b)=y_i$ be the (unit speed) line segment running from $x_i$ to $y_i$ parallel to $e_1$. Since $(F_1\circ\gamma)'(t)>0$ for all $t$ and $s(\gamma(t))<\epsilon\implies|(G\circ\gamma)'(t)|<\epsilon|(F_1\circ\gamma)'(t)|$,
$$|G(y_i)-G(x_i)|\leq\int_a^b|(G\circ\gamma)'(t)|dt<\epsilon\int_a^b|(F_1\circ\gamma)'(t)|dt=\epsilon\,|F_1(y_i)-F_1(x_i)|.$$
