I want to proof, that
$X=\{ (p,q)\in S^n\times S^n,\ p\neq q \}$ is of the same homotopy-type as the $S^n$.
By taking the "straight line homotopy" for fixed $a\in S^n$ in the following form
$R(p,q,t)=\frac{t(p,q)+(1-t)(p,a)}{\|t(p,q)+(1-t)(p,a)\|}$,
$R$ is a deformation from $S^n\times S^n\rightarrow S^n$, because
$R(p,q,0)\in S^n,\ R(p,q,1)=(p,q),\ R(p,a,t)=(p,a).$
Hence $S^n$ is a deformation retract, $R$ composed with the inclusion map
$i:S^n\rightarrow S^n\times S^n$, it is homotopic to the identity map on $S^n\times S^n$. The homotopy aquivalence follows immediately.
Because I am new to the idea of homotopy, could you please read that sketch of proof and tell me if it's right or not?
Thank you in advance!
It is not true. This is most obvious for $n = 0$ even without using algebraic topology.
Edited:
The original question was whether $S^n \times S^n$ has the same homotopy type as $S^n$. This is false for all $n$. The corrected question is whether $X = S^n \times S^n \setminus D$, where $D = \{(p,q) \in S^n \times S^n \mid p = q \}$ is the diagonal, has the same homotopy type as $S^n$.
This is true. First we shall specify an embedding $i : S^n \to X$. We cannot take $i(p) = (p,a)$ with some fixed $a \in S^n$ because $(a,a) \notin X$. However, we may take $$i(p) = (p,-p) .$$ This is a well-defined continuous injection (note $(p,-p) \notin D$) and therefore an embedding because $S^n$ is compact. This gives us a homeomorphism $\bar i :S^n \to \bar S^n = i(S^n)$. We next show that $\bar S^n$ is a strong deformation retract of $X$. Define $$R : X \times I \to X, R(p,q,t) = \left(p,\dfrac{t(-p) + (1-t)q}{\lVert t(-p) + (1-t)q \rVert} \right).$$ Let us check for which $(p,q,t) \in S^n \times S^n \times I$ we have $t(-p) + (1-t)q = 0$. It is impossible for $t = 0$ because $q \ne 0$. Hence we get $p = (\frac{1}{t} - 1)q$ which implies $1 = \lVert p \rVert = \lvert \frac{1}{t} - 1 \rvert \cdot \lVert q \rVert = \lvert \frac{1}{t} - 1 \rvert$, i.e. $\frac{1}{t} - 1 = 1$ which means $t = \frac{1}{2}$ or $\frac{1}{t} - 1 = -1$ which does not have a solution. Inserting $t = \frac{1}{2}$ yields $p = q$, i.e. $(p,q) \in D$.
Therefore, if $(p,q,t) \in X \times I$, then $t(-p) + (1-t)q \ne 0$. Let us next verify that on $X \times I$ it is impossible that $\frac{t(-p) + (1-t)q}{\lVert t(-p) + (1-t)q \rVert} = p$ which would mean $R(p,q,t) \in D$. Assume that $t(-p) + (1-t)q = rp$ with $r = \lVert t(-p) + (1-t)q \rVert > 0$. Then $\frac{1-t}{r+t}q = p$. Taking the norm on both sides yields $\lvert \frac{1-t}{r+t}\rvert = 1$. Since $\frac{1-t}{r+t} \ge 0$ this implies $\frac{1-t}{r+t} = 1$, hence $q = p$ which is impossible for $(p,q) \in X$.
Thus $R$ is well-defined. We have $R(p,q,0) = (p,q)$, $R(p,q,1) = (p,-p) \in \bar S^n$ and $R(p,-p,t) = (p,-p)$. This means that $\bar S^n$ is a strong strong deformation retract of $X$.