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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!

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    $\begingroup$ They are not of the same type, and in your proof $R$ is ill-defined (sometimes you divide by 0). $\endgroup$ – Michal Adamaszek Sep 2 '19 at 11:37
  • $\begingroup$ You should say $n$-sphere instead of $n$-circle. $\endgroup$ – Paul Frost Sep 2 '19 at 11:38
  • $\begingroup$ I forgot to mention, that $p\neq q$. The original Set is $X=\{(p,q)\in S^n\times S^n ,\ p\neq q\}$ $\endgroup$ – whiteian motion Sep 2 '19 at 13:27
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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 $X$.

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  • $\begingroup$ I truly appreciate your answer! $\endgroup$ – whiteian motion Sep 2 '19 at 17:22

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