How to show that $S^n$ can't embed $\mathbb R^n$ There are some hints that we can prove it by the homology group of $S^n\setminus S^k$.
 A: The easiest way is by point set topology plus invariance of domain. The sphere is compact, so it’s image under an embedding would be closed. However, it is locally homeomorphic to $\mathbb{R}^n$ so it must be open. Since the only non empty closed and open subset of $\mathbb{R}^n$ Is itself, this must be a bijective embedding. However, this is impossible since $\mathbb{R}^n$ is not compact. Hence, no embedding can exist.
A: If you want to use homology, you can use the fact that if $f:S^{n-1}\to S^n$ is an embedding, then $\tilde H_0(S^n-f(S^{n-1}))=\mathbb Z.$
Now suppose that $S^n$ embeds in $R^n$. Then, since $\mathbb R^n$ embeds in $S^n$ non-surjectively, we get an embedding $f:S^n\to S^n$ which is not surjective.
Now, Mayer-Vietoris now gives the exact sequence:
$\tilde H_0(S^n-f(S^n_+)\oplus\tilde H_0(S^n-f(S^n_-)\to H_0(S^n-f(S^{n-1}))\to 0.$
But the left group is $0$, while the middle one is $\mathbb Z$, which implies that $(S^n-f(S^n_+))\cap (S^n-f(S^n_-))=\emptyset$ and so $f(S^n)=S^n$, a contradiction.
