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.
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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
- $a = a' = 1$ or
- $a = a' = -1$, or
- $a = a'$ 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.
John Hughes
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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.
SvanN
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Ronnie Brown
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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)!
Daniel Valenzuela
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thank you. Ok, you mean "=" is not true in general and the definition on the left site of the "equality" ("=" ) is the standarddefinition? – Oct 05 '14 at 14:58
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Yes precisely! The LHS is equivalent to the standard definition (with $X\times 1$ I mean the image of this subset in the quotient) and explains why many people abuse the notation. – Daniel Valenzuela Oct 05 '14 at 15:15
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ok, thank you, I understand. – Oct 05 '14 at 15:20