# Deformation Retract of subset of $S^n\times S^n$

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.

• $H$ is into $X$. Jun 7, 2023 at 22:28

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 $$\in I$$.