The book can be found here.
The exercise asks the following:
Show that $\mathbb{R}^n$ does not contain any subset homeomorphic to $D^m$ with $m>n$, where $D^m$ is the $m$-dimensional ball.
First of all, if I call $A\subseteq\mathbb{R}^n$ a set which I'm going to assume is homeomorphic to $D^m$ via $\phi$, then $A$ is compact and connected. In particular $A$ is closed.
The case when $n=1$ is easy, because the connected compact subsets of $\mathbb{R}$ are the closed and bounded intervals, which are easily shown not to be homeomorphic to $D^m$ for $m>1$.
For the general case, from lemma 7.6, I know that there exists a homeomorphism $h:\mathbb{R}^{n+m}\to\mathbb{R}^{n+m}$ such that $h(x,0_m)=(0_n,\phi(x))$ for all $x\in A$. Hence, I can consider $\mathbb{R}^{n+m}-A$ identifying $A$ with $A\times \{0_m\}$ and $\mathbb{R}^{n+m}-D^m$ identifying $D^m$ with $h(A, 0_m)=(0_n,\phi(A))=(0_n,D^m)$. And what's more, these sets are homeomorphic.
From this point, I was thinking about using the cohomology of this two subspaces, which should be identical, and come to a contradiction, but I can't find the way.
Is this a right way to face this problem? In that case, how could I continue? If not, what would be a more straightforward way to solve it? I also thought about using invariance of domain, but I didn't know how.
Note: I'm only supposed to use the deRham cohomology of open euclidean spaces.
It's a lot easier. If $A$ is homeomorphic to $D^m$ then their interiors are homeomorphic too. Hence, one would have an open set in $\mathbb{R}^n$ (the interior of $A$) homeomorphic to an open set in $\mathbb{R}^m$ (the interior of the ball), which is not possible by the invariance of domain.