I think you should tag your questions with "homework" if they are homework, as Zhen Lin pointed you out.
Anyway, here is my solution. It just needs a little bit of notation. Write the points of $\mathbb{R}^{n+1} = \mathbb{R}^{m+1} \times \mathbb{R}^{n-m}$ as $(x,y)$, with $x\in\mathbb{R}^{m+1}$ and $y \in \mathbb{R}^{n-m}$.
So, $S^m$ is embedded in $S^n \subset \mathbb{R}^{n+1}$ as those $(x,y) \in\mathbb{R}^{n+1} $ such that $y=0$ and $\| x \| = 1$. Likewise, embed $S^{n-m-1}$ in $S^n$ as those $(x,y)$ such that $x=0$ and $\|y\| =1$. So, $S^{n} \backslash S^m$ are those $(x,y)\in S^n$ such that $y\neq 0$. Briefly:
- $S^m = \left\{ (x,y) \in \mathbb{R}^{n+1} \ \vert \ y = 0 \ , \ \|x \|=1 \right\}$
- $S^{n-m-1} = \left\{ (x,y) \in \mathbb{R}^{n+1} \ \vert \ x = 0 \ ,\ \|y \|=1 \right\}$
- $S^n \backslash S^m = \left\{ (x,y) \in \mathbb{R}^{n+1}\ \vert \ y \neq 0 \ , \ \|x\|^2+\|y\|^2 =1 \right\} $
Then, our inclusion is just
$ i : S^{n-m-1} \longrightarrow S^n\backslash S^m \ , \qquad i(0,y) = (0,y) \ . $
In the other direction, now the map is pretty obvious: since points $(x,y) \in S^{n-m-1}$ are those with $x = 0$, given a point $(x,y) \in S^n\backslash S^m$, the simplest way to obtain a point in $S^{n-m-1}$ seems to put $x=0$ and then normalize the second coordinate:
$ f: S^n \backslash S^m \longrightarrow S^{n-m-1} \ , \qquad f(x,y) = \left( 0,\frac{y}{\|y\|} \right) \ . $
Notice that, since $y\neq 0$, $f$ is continuous and
$ (f\circ i) (0,y) = f(0,y) = \left( 0,\frac{y}{\|y\|} \right) = (0,y) \ , $
because $\|y\|=1$.
As for the other composition, we have
$ (i\circ f)(x,y) = \left( 0,\frac{y}{\|y\|} \right) \ , $
and we need now an homotopy between $\mathrm{id_{S^n\backslash S^m}}$ and $i\circ f$. For this, the old trick of the straight line homotopy normalized suffices. Define
$ H : (S^n\backslash S^m)\times I \longrightarrow S^n\backslash S^m \ , \qquad H(x,y,t) = \frac{\left((1-t)x, (1-t)y + t\frac{y}{\|y\|} \right)}{\left\| \left((1-t)x, (1-t)y + t\frac{y}{\|y\|} \right) \right\|} \ . $
And we have
$ H(x,y,0) = \frac{( x,y )}{\|(x,y)\|} = (x,y) \ , $
since $(x,y) \in S^n$ and
$ H(x,y,1) = \frac{\left(0, \frac{y}{\|y\|} \right)}{\left\| \left(0, \frac{y}{\|y\|} \right) \right\|} = \left( 0, \frac{y}{\|y\|} \right) \ . $
We need to prove two more things:
- That $H(x,y,t) \in S^n\backslash S^m$ for all $(x,y,t) \in (S^n\backslash S^m)\times I$.
- And that $H$ is continuous.
We can do it in one go, just showing that $(1-t)y + t\frac{y}{\|y\|} \neq 0 $ for all $(x,y,t) \in (S^n\backslash S^m)\times I$. But
$ (1-t)y + t\frac{y}{\|y\|} = 0 \qquad \Longleftrightarrow \qquad (1-t)y\|y\| +ty = 0 $
Since $y\neq 0$, this would be the same as
$ (1-t)\|y\| +t = 0 \ . $
But, for $t \in [0,1]$, the left-hand side of the above equation represents points in the segment between $\|y\|$ and $1$. Since $y \neq 0$, this segment does not include the point $0$. Hence we are done.