# The n-Sphere is of the same homotopy-type as the product of 2 “almost” n-Spheres

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?

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

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 $$X$$.

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