Exercise $2.1.1$ - Differential Topology by Guillemin and Pollack If $U \subset \mathbb{R}^k$ and $V \subset H^k$ are neighborhoods of $0$, prove that there exists no diffeomorphism of V with U. 
Here, $H^k$ is simply the upper half-space.
I tried to solve this problem with the continuity, but it is a dead-end. I have difficulty to solve this problem. Anyone could give me a hint to complete this problem?
 A: Let $\ f:U \to V$ be such  a diffeomorphism. Let $x=f^{-1}(0)$. For $h$ near $x$ we have $f(h)=0+Df_x(h-x)+e(h-x)$ where $\frac {e(h-x)}{|h-x|} \to 0$ as $|h-x| \to 0$. As $f$ is a diffeomorphism, $Df_x$ is a linear isomorphism (since $D(f^{-1})_{f(x)}=(Df_x)^{-1}$ by differentiating $ f^{-1}\circ f =I$ using the chain rule). 
Let $V=(Df_x)^{-1}(-1,0,0,\dots,0)$, and $h=x+\frac1NV $ with $N$ so that $h\in U$ and $$\frac{|e(\frac1NV)|}{|\frac1NV|}<\frac1{|V|}$$
Now $f(h)=0+(-\frac{1}N,0,0,\dots,0)+e(\frac1NV)$. As $|e(\frac1NV)|<\frac1N$, $f(h)$ is not in $\Bbb H^k$. So no such diffeomorphism could exist.
A: Some of the comments have suggested using algebraic topology.  That sounds horrifying.  Let's be analysts and ignore hard stuff.  The idea is pretty clear.  If $V$ is a neighborhood of $0$ in $H^k$, it's going to have to contain a piece of the boundary.  But there's nowhere for the boundary to map to in $U$.
If $V$ is a neighborhood of $0$ in $H^k$, it contains a neighborhood of the form $D = \{x = (x_1, ..., x_k): |x| < \epsilon, x_k \ge 0\}$ for some $\epsilon > 0$.  Let $f:U \to V$ be the supposed diffeomorphism.  Pick $x \in U$ so that $f(x) \in D$.
Since $f$ is differentiable at $x$, we can write $f(s) = f(x) + Df_x(s-x) + h(s-x)$, where $\frac{|h(s-x)|}{|s -x|} \to 0$ as $s \to x$.  Since $f$ is actually a diffeomorphism, $Df_x$ is actually a linear isomorphism.  
At this point, we'd like to force $s$ out of $H^k$.  Choosing a new coordinate system if necessary, we may take $Df_x$ to be the identity map, so really near $x$ we have that 
A brief aside to justify the above: Let $V$ be a finite dimensional real vector space of dimension $k$.  Let $T:V \to V$ be a linear isomorphism.  Choose any basis $v_1, ..., v_k$ for $V$.  Let $w_j := T^{-1}(v_j)$.  Then $w_1, ... , w_k$ is also a basis for $V$ (you should check this).
Now let's think of $T:(V, \{w_1, ... , w_k\}) \to (V, \{v_1, ..., v_k\})$.  What does $Tx$ look like for $x \in V$?  Since $T$ has the property that $Tw_j = v_j$ for any $j$, if $x = x_1 w_1 + \dots + x_k w_k$, then we have that $Tx = x_1 v_1 + \dots + x_k v_k$.  This is what I mean by representing $T$ by the identity through a choice of basis.
Going back to the above, we've written $f(s) = f(x) + Df_x(s-x) + h(s-x)$ for $Df_x$ a linear isomorphism from $\mathbb{R}^k \to \mathbb{R}^k$.  Choose a basis $v_1, \dots, v_k$ for $\mathbb{R}^k$ so that $Df_x(v_j) = e_j$, where $e_j$ is the $j$th basis vector.
Let $| \cdot |_{v}$ denote the norm on $\mathbb{R}^k$ with respect to this new basis.  Since $\mathbb{R}^k$ is a finite dimensional vector space, all norms are equivalent (exercise!).  Choose a constant $c > 0$ so that $ c^{-1} |\cdot| \le | \cdot|_{v} \le c |\cdot|$.
Since $x \in U$ and $U$ is open, choose $\eta > 0$ so that $B_v(x, \eta) \subset U$, where the ball is taken in our new norm (i.e., $B_v(x, \eta) = \{y \in \mathbb{R}^k: |x - y|_v < \eta\}$).  
Choose $s = x - \frac{\eta}{2} v_k$.  Clearly, $s \in B_v(x, \eta)$.  
Shrinking $\eta$ if necessary, we will have that $|h(s-x)| \le \frac{1}{2}|s-x|$ so long as $|s-x| < \eta$ as well.  
So, $f(s) = f(x) + Df_x(s-x) + h(s-x) = f(x) + Df_x(x - \frac{\eta}{2}v_k -x) + h(s-x)$
$= f(x) - \frac{\eta}{2}e_k + h(s-x)$.
Now $f(x) \cdot e_k = 0$, by assumption (we chose $x$ specifically so that the $k$th component of $f(x)$ is $0$).  Additionally, the above bound on $h(s-x)$ implies that $h(s-x) \cdot e_k \le \frac{\eta}{4}$.
Together, this shows that $f(s) \cdot e_k = -\frac{\eta}{2} + h(s-x) \cdot e_k \le -\frac{\eta}{4} < 0$, which implies that $f(s) \notin V$; this is a contradiction.
A: Here's the horrible algebraic topology proof that there cannot even be any homeomorphism $f : U \to V$.
First, for every open $U' \subset U$ and every $x \in U'$ we have the following calculation of relative homology:
$$H_k(U',U'-x) \approx \mathbb{Z}
$$
To see why, one chooses an open ball $B \subset U'$ such that $x \in B$, and then one has isomorphisms
$$H_k(U',U'-x) \approx H_k(B,B-x) \approx \mathbb{Z}
$$
The first of these two isomorphisms is an application of the excision theorem, the second is one of the basic steps in the calculation of the homology of spheres.
Second, there exists an open $V' \in V$ and $y \in V'$ and such that
$$H_k(V',V'-y) \approx 0
$$
To see why, one chooses $y$ in the boundary of $V$ and then one chooses $V'$ to be a half-ball whose boundary contains $y$. Then one applies the long exact sequence of the homology of a pair, which has the following consecutive terms:
$$H_k(V') \mapsto H_k(V',V'-y) \mapsto H_{k-1}(V'-y)
$$
The first and last terms are zero because $V'$ and $V'-y$ are contractible (here I assume $k \ge 2$, alternatively if $k=1$ use reduced $H_0(V'-y)$ which will be zero). Therefore the middle term is zero by exactness.
Finally, if there existed a homeomorphism $f : U \to V$ then we could choose $y \in V' \subset V$ as above so that $H_k(V',V'-y) \approx 0$, and then we could set $U' = f^{-1}(V')$ and $x = f^{-1}(y)$, and one would obtain $H_k(U',U'-x) \approx \mathbb{Z}$. But, the homeomorphism $f : (U',x) \mapsto (V',y)$ induces an isomorphism 
$$f_* : \mathbb{Z} \approx H_k(U',U'-x) \to H_k(V',V'-y) \approx 0
$$
which is a contradiction.
A: Define the tangent cone $C_p$ at a point $p$ to be the subset of $T_pM$ given by $\gamma'(0)$, where $\gamma: [0,\varepsilon) \to M$ is a smooth curve. At a boundary point $p \in \mathbb H^k$ I claim $C_p$ is the closed upper-half space. It obviously includes that; but on the other hand given $\gamma: [0,\varepsilon) \to \mathbb H^k$, we have $\gamma(t) = \int_0^t \gamma'(s)ds$, so if $\gamma'(0)$ is not in the upper-half space, neither is $\gamma(t)$ for sufficiently small $t$.
Because diffeomorphisms would preserve tangent cones, and the tangent cone of any point on the interior is $T_p M$, there is no desired diffeomorphism.
This argument generalizes to showing that manifolds with various kinds of corners (squares, pyramids of various shapes, etc) are not diffeomorphic to other kinds: you just show there is no linear isomorphism between the tangent spaces that preserves the tangent cones.
A: Here are two possibilities:
Argument 1
Suppose $f$ was a diffeomorphism from $U \subseteq \mathbb{R}^k$ to $V \subseteq \mathbb{H}^k$, and that $f(p)=0$. By shrinking $V$ and $U$, we may assume that $V$ is of the form
$$V=\{x\in \mathbb{H}^k\;| \; \| x \|_2<\varepsilon\},$$
for some $\varepsilon>0$. Thus, we may also assume that $U$ is connected, hence path connected (being an open subset of $\mathbb{R}^k$). Under these assumptions, $f$ must also be a diffeomorphism of $U \setminus \{p\}$ with $V \setminus \{0\}$.
Claim:
Any smooth map $\varphi:S^{k-1} \to V\setminus \{0\}$ is smoothly homotopic to a constant map.
Proof: Let $x_0=(0,0,\ldots,0,\varepsilon/2)$. Then the map $F:S^{k-1}\times[0,1]\to V\setminus \{0\}$ given by
$$F(s,t)=tx_0+(1-t)\varphi(s)$$ is clearly smooth, has
$F(s,0)=\varphi(s)$ and $ F(s,1)=x_0$ for all $s \in S^{k-1}$, and maps into $V\setminus \{0\}$, since it is a straght line homotopy. $\square$
Low let us show that $U \setminus \{p\}$ does not have this property. Since $U$ (with $p$ included) is an open set, we may find an embedding $\phi$ of $S^{k-1}$ onto
$$K=\{x \in U \; | \;\; \| x-p\|_2=r\}$$
for some $r>0$. Let $\gamma:K\to U\setminus\{p\}$ be the antipodal map of $K$, i.e. $\gamma(q)=p+2(p-q)=3p-2q$. by the Borsuk-Ulam theorem, the mod 2 winding number $W_2(\gamma, p)=1$.
Suppose now $F$ was a smooth homotopy of $\phi$ to a constant map $g:s\to x_0$. We may assume that $x_0 \in K$. Then this would give rise to a smooth homotopy of $\gamma$ with the constant map $\Gamma:K \to U\setminus \{p\}, \Gamma (k)=x_0$. Since the mod 2 winding number is invariant under smooth homotopies, we would get
$$1=W_2(\gamma,p)=W_2(\Gamma,p)=0,$$
a contradiction.
Argument 2 (although not super rigorous.)
Assume that we have argued from the assumptions that the derivative $df_x$ must be a linear isomorphism for at every $x \in U$. V must contain a segment
$$r_\varepsilon=\{(0,0,\ldots,0,t)\;| \;0\le t\le \varepsilon\}$$
for some $\varepsilon >0$. Consider the preimage $\mathcal{\gamma}$ of $r_\varepsilon$ in $U$. This must be a smooth curve with endpoints $p \ne q$. Assume that $f(p)=0$. Now extend $\gamma$ to a smooth curve $\Gamma$ at $p$ in $U$, such that $\Gamma$ has $p$ as an interior point. We may assume that $\Gamma$ and $\gamma$
is parametrized so that $$\Gamma:[-\delta,\varepsilon]\to U,$$ where $\delta>0$, $\Gamma(-\delta)=r,\Gamma(t)=\gamma(t)$ for $t \in [0,\varepsilon]$. Now attempt to compute
$$d(f\circ \Gamma)_0(t)=df_p \big (t\Gamma'(0) \big )$$
for $t \ne 0$. We may assume that $\Gamma'(0)\ne 0$. Now $f(\Gamma(t)) \in \mathbb{H}^k$ for arbitrary $-\delta<t<0$, so $f(\Gamma)$ must have a cusp at zero. This is only possible if the derivative of $(f\circ \Gamma)'(0)$ is zero or undefined. Since it is defined by assumption, we are forced to conclude (setting $t=1$) that
$$df_p(\Gamma'(0))=0.$$
Since $\Gamma'(0)\ne 0$ by assumption, we get that $df_p$ has a nontrivial kernel, so $df_p$ cannot be a linear isomorphism, a contradiction.
