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I have been given the following excercise: Let $X\subset S^n\times S^n, \ n\geq 1$ defined as $X=\{(p,q) \in S^n \times S^n; p\neq q\}$.

Show that $Y=\{(p,-p); p \in S^n\}$ is a strong deformation retract.

Help: Build the Homotopy as $H:X\times I \longrightarrow Y$ as follows:

$$H(p,q,t) = (p,G(p,q,t))$$ Using that $(1-t)q-tp\neq 0 \ \forall t \in [0,1], \ \forall p\neq q$

Mi idea was to focus in one $p$ in particular, since for a fixed $p$, the points in the space are $S^n-\{p\}$ which is homemorphic to the plane and therefore there exists a Homotopy relative to $\{-p\}$ for each $p$: $$G_p(q,t) : (S^n-\{p\})\times I \longrightarrow S^n-\{p\}$$ such that $G_p(q,0)=q$ and $G(q,1) = -p$.

Then my idea was to use $H(p,q,t) = (p,G_p(q,t))$ as a homotopy as $H(p,q,0) =(p,q)$ and $H(p,q,1)=(p,-p)$

However I shold be able to show that $H$ is continuous, and I don't know if I can say that the union of these functions is continuous as separately they are continuous.

Also I haven't used the secod part of the "help" comment, so I don't know if this is the way to go.

Thanks for any help.

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  • $\begingroup$ $H$ is into $X$. $\endgroup$
    – Bob Dobbs
    Jun 7, 2023 at 22:28

1 Answer 1

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The idea is nice, but the problem is that you do not specify what $G_p$ is exactly. Thus you are not able to prove the continuity of $H$. Here is a suggestion. Define $$G : X \times I \to S^n, G(p,q,t) = \frac{-tp+(1-t)q}{\lVert -tp+(1-t)q\rVert}, $$

$$H : X \times I \to X, H(p,q,t) = \left(p, G(p,q,t) \right).$$ To show that these functions are well-defined we use the following

Lemma. Let $p, q \in S^n$ and $p = \lambda q$ for some $\lambda > 0$. Then $\lambda = 1$, i.e. $p = q$.

Proof. We have $1 = \lVert p \rVert = \lVert \lambda q \rVert = \lambda \lVert q \rVert = \lambda$.

  1. We have $-tp+(1-t)q \ne 0$ for all $t \in I$, thus $G$ is well-defined.

This is obvious for $t = 0,1$. Assume that $-tp+(1-t)q = 0$ for some $t \in (0,1)$. Then $p= \frac{1-t}{t}q$. Since $\frac{1-t}{t} > 0$, we get $p = q$ which contradicts $(p,q) \in X$.

  1. We have $p \ne G(p,q,t)$ for all $t \in I$, i.e. $H(p,q,t) \in X$. Thus $H$ is well-defined.

This is again obvious for $t = 0,1$. Assume that $p = G(p,q,t)$ for some $t \in (0,1)$. This implies $p = \frac{1-t}{t + \lVert -tp+(1-t)q\rVert}q$ which again implies $p = q$.

By definition $H(p,q,0) = (p,q)$, $H(p,q,1) = (p,-p)$ and $H(p,-p,t) =(p,-p)$ for all $ \in I$.

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