unreduced suspension

Question

Is the definition $SX=\frac{(X\times [a,b])}{(X\times\{a\}\cup X\times \{b\})}$ of the unreduced suspension the standard defininition? If I consider $X=$ point, the suspension of $X$ is a circle. But I saw an other definition of the unreduced suspension such that the suspension of a point should be an interval. Regards.

Answer 1

I would have said that the suspension was $X \times [-1, 1]$ modulo the relation that $$ (x, a) \sim (x', a') $$ if and only iff

1. $a = a' = 1$ or
2. $a = a' = -1$, or
3. $a = a'$ and $x = x'$.

Wikipedia seems to agree with me. It looks as if your author was a little glib, and failed to mention that the "bottom" and "top" sets of equivalent points were not supposed to be made equivalent to each other.

Answer 2

In fact in my experience the unreduced suspension is commonly written in the way you did it, although it is wrong. What it means is: collapse one side to a point and also the other side. But NOT both to the same point. So this would be only right when quotiening is associative, i.e. $$ SX = (X \times I/ X\times \{0\} )/ (X \times \{1\}) "=" X \times I / X\times 0 \cup X\times 1 $$

To answer your question clearly: no this definition is wrong (or at least not standard)!

Answer 3

One should have a picture and here is one, taken from the e-version of Topology and Groupoids showing the suspension $SX$ as a union of two cones:

Note the special case when $X$ is a circle $S^1$ when this gives the $2$-sphere $S^2$ as the union of two hemispheres.