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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeMay 31st 2011
• (edited Nov 4th 2013)

quick fix at suspension: distinction between plain and reduced/based suspension. More should be said here, but not by me right now.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeMay 31st 2011
• (edited May 31st 2011)

But reduced suspension already has its own page (started by Tim Porter), which has been linked from suspension for some time. I’ve edited these pages and suspension object accordingly. (Please confirm that Tim’s reduced suspension is the same as yours.)

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeMay 31st 2011

Thanks, Toby. I didn’t see this for some reason.

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeJun 1st 2011

I wrote:

Please confirm that Tim’s reduced suspension is the same as yours.

Well, you both said that it’s the smash product with the circle, so that should be fine (although you seem to care about it only up to homotopy).

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeJun 1st 2011
• (edited Jun 1st 2011)

Sorry Toby, I had forgotten to reply to your request:

yes, it seems what Tim had written is the standard construction, And, yes, I think that the concept of suspension is not too interesting up to homeomorphism, but natural in a context up to weak homotopy equivalence.

On all these points the entry deserves further expansion. But not by me right now.

• CommentRowNumber6.
• CommentAuthorzskoda
• CommentTimeJun 1st 2011

Urs, I could agree about strong homotopy equivalence, not weak one. The suspension is important in shape theory for example, where going up to weak equivalence would be a disaster for some spaces.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeJun 1st 2011

Okay, I don’t really know about that. So maybe I should say strong homotopy equivalence .

• CommentRowNumber8.
• CommentAuthorTobyBartels
• CommentTimeJun 1st 2011

The suspension is useful up to homeomorphism, where it defines the continuous structure of spheres, and even up to diffeomorphism, where it defines the smooth structure of cubes. See the examples at suspension. (It seems that the spell checker in Firefox 4 knows “homeomorphism”, but not “diffeomorphism” yet).

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeJun 2nd 2011

The suspension is useful up to homeomorphism, where it defines the continuous structure of spheres, and even up to diffeomorphism, where it defines the smooth structure of cubes.

Okay, but doesn’t the very term “suspension” for the construction that the explicit formulas are models for refer to the homotopy-theoretic interpretation? Doesn’t it refer to the shifting-up of homotopy groups? (I don’t actually know, maybe not, but that’s what I always thought.)

• CommentRowNumber10.
• CommentAuthorAndrew Stacey
• CommentTimeJun 2nd 2011

I think you mean “homology”. Suspension shifts homology but is more complicated on homotopy.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeJun 2nd 2011

Right, I misspoke. Sorry.

• CommentRowNumber12.
• CommentAuthorjim_stasheff
• CommentTimeJun 2nd 2011
doesn't the very term "suspension" for the construction that the explicit formulas are models for refer to the homotopy-theoretic interpretation?

I would say NO that would be revisionist history
initially I think the term `suspension' was motivated by the picture
the precise construction was useful in several ways

it then spawn adjectival versions such as the suspension homomorphism in both homology and homotopy
• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeJun 2nd 2011

Okay, I see.

• CommentRowNumber14.
• CommentTimeMar 25th 2014
• (edited Mar 25th 2014)

Let C be a pointed symmetric monoidal model category. One can define suspension as the homotopy colimit of the diagram $\ast \leftarrow X \to \ast$, as is done at suspension object, or as $X \otimes S^1$, where $S^1$ is the suspension of $S^0 = \ast \sqcup \ast$ in the former sense. Are these the same?

Maybe this only works in the case where C is the category of pointed objects in some cartesian model category, with the usual model structure and the smash product. In that case I think I have a proof, but it seems too easy…

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeMar 25th 2014

So if the tensor is indeed the smash product and preserves hocolims, then it just follows directly, that’s probably what you have in mind.