Show that an inclusion is an isomorphism in homology I'm struggling a bit with an exercise from a book, in a chapter about the Jordan-Brouwer separation theorem. It goes as follows:
(note: $s_{n-1}$ is a topological space homeomorphic to $\mathcal{S}^{n-1}$)
Let $x$ be a point in the bounded component of $\mathbb{R}^n-s_{n-1}$, $n\ge 1$. Show the inclusion $s_{n-1}\to \mathbb{R}^n-\{x\}$ induces an isomorphism in homology. Conversely, if $x$ lies in the umbounded component, the inclusion induces the zero map. Suggestion: consider deformations of $\mathbb{R}^n-\{x\}$ to spheres of constant radius and center $x$.
Thank you very much for your help!
 A: First, you want to show that $\mathbb{R}^n \setminus\{x\}$ deformation retracts to an embedded $S^{n-1}$ centered at $x$. This is fairly easy to see intuitively for $n = 1,2,3$, and can be proven in general by drawing a line from $y$ to $x$ for any $y \neq x$, and then if $d(y,x) < 1$ sliding $y$ out to a point on the line distance $1$ from $x$, and if $d(y,x) > 1$, sliding inward. This can be written explicitly as a homotopy in coordinates by taking a linear homotopy $t y + (1-t)y'$, where $y'$ is the intersection of the ray from $x$ to $y$ with the sphere of radius $1$ around $x$. 
Now, under this homotopy, the inclusion of an embedded $s_{n-1}$ such that $x$ is in the bounded component of the complement will be taken to the sphere (if $x$ is not in the bounded component, it would be taken to some contractible patch on the sphere), so the inclusion from $s_{n-1}$ to $\mathbb{R}^n\setminus\{x\}$ is homotopic to a homeomorphism (which might be orientation-reversing) $S^{n-1} \rightarrow S^{n-1}$, so it induces an isomorphism on homology.
