# Suspension of the $(-1)$-sphere

I have a doubt about the construction of the suspension of the sphere of dimension -1, that is the empty set. In the following image I try to explain the suspension of the empty set: My doubt is that I think that the "part" of the empty set that result of the product of one of the endpoints of the line with the empty space is collapsed with a point and we proceed in this way with the other remaining point, but the empty space no have points, then, where the points come from? Am I correct with the thinking of collapse the empty space with a point "two times"?

• I have no sensible way of answering your question so I will just say that perhaps sometimes definitions shouldn't be extended to pathological cases. – Jacob Bell May 14 '12 at 17:45

(Reduced) suspension is equivalent to taking $S^{0} * X$, where $*$ is the (reduced) join and $S^{0}$ is the $0$-sphere. If $X$ is empty, then $\Sigma X \simeq S^{0}$, which is precisely the two-pointed space that you compute.
• Sure, if $X$ is not empty. – user02138 May 23 '12 at 21:20
The whole question is maybe somewhat up to convention. But the usual convention makes sense here. If we look at the definition of the quotient for spaces $A\subset X$ carfully, then we see that $X/A$ consist of one point $[A]$ and all the points of $X-A$. If we now consider $X/\emptyset$, then this is all the points of $X$ and one point $[\emptyset]$. Hence if you take the unreduced suspension you end up with two points.
• And that is good as the discrete 2 point space is the $0$-sphere, being the boundary of the 1-ball, which is a closed interval! – Ronnie Brown May 15 '12 at 17:05