Deformation Retractions

peccavi_2006 · **Posted:** Fri, 7 Oct 2011 20:20:23 UTC

Today in lectures we were working on deformation retractions and did the standard examples ( $\mathbb{R}$ deformation retracts to a point, and $\mathbb{R}^{n} - \{0\}$ deformation retracts to $S^{n-1}$ ), but I was wondering about tori.

I know the 2-torus is the product of two circles, so would I be right in assuming that the 2-torus deformation retracts to the wedge of two circles?
And then, if I removed a point on the torus, how would that affect it? Because obviously I would no longer have two circles because one of the circles wouldn't be closed anymore...

Thank you!

Shadow · **Posted:** Fri, 7 Oct 2011 21:51:36 UTC

peccavi_2006 wrote:

Today in lectures we were working on deformation retractions and did the standard examples ( $\mathbb{R}$ deformation retracts to a point, and $\mathbb{R}^{n} - \{0\}$ deformation retracts to $S^{n-1}$ ), but I was wondering about tori.

I know the 2-torus is the product of two circles, so would I be right in assuming that the 2-torus deformation retracts to the wedge of two circles?
And then, if I removed a point on the torus, how would that affect it? Because obviously I would no longer have two circles because one of the circles wouldn't be closed anymore...

Thank you!

No, that's impossible, a deformation retract is a homotopy equivalence, and it's easy to see those two spaces do not have the same homotopy type, since homology is a homotopy invariant.

peccavi_2006 · **Posted:** Sat, 8 Oct 2011 22:38:26 UTC

Ah, we haven't done homology yet

If I considered the torus as the quotient of a square in the usual way (opposite sides identified with the same orientations), and puncture a hole in it, then can I (forgoing any sort of use of actual terminology) expand the hole so that I only have the boundary of the square left? And then identifying the edges I get the wedge of two circles?
Does that sound more...usual?

And then, would a simple circle be the deformation retraction of the torus?

But then, if I had a surface of genus g, $\Sigma_{g}$ I could represent that as the connected sum of g tori...but then I don't know how to go about using the above approach to determine what it retracts to.
I mean, if I removed one point, I must still have the figure of 8 graph in there somewhere, but what do I do with the other (g-1)

tori that don't have a hole removed?

Sorry if this is all total garbage again - I haven't quite got my head around it yet

Shadow · **Posted:** Sat, 8 Oct 2011 22:44:45 UTC

peccavi_2006 wrote:

Ah, we haven't done homology yet

If I considered the torus as the quotient of a square in the usual way (opposite sides identified with the same orientations), and puncture a hole in it, then can I (forgoing any sort of use of actual terminology) expand the hole so that I only have the boundary of the square left? And then identifying the edges I get the wedge of two circles?
Does that sound more...usual?

And then, would a simple circle be the deformation retraction of the torus?

But then, if I had a surface of genus g, $\Sigma_{g}$ I could represent that as the connected sum of g tori...but then I don't know how to go about using the above approach to determine what it retracts to.
I mean, if I removed one point, I must still have the figure of 8 graph in there somewhere, but what do I do with the other (g-1)

tori that don't have a hole removed?

Sorry if this is all total garbage again - I haven't quite got my head around it yet

Do you at least know about the fundamental group? The torus cannot be a deformation retract of the circle because those induce isomorphisms on the fundamental group, but the circle's fundamental group is $\mathbb{Z}$ and the torus' is not.

peccavi_2006 · **Posted:** Sun, 9 Oct 2011 00:00:20 UTC

Oh yeah - $\pi_{1} (S^{1}) = \mathbb{Z}$ , whereas $\pi_{1} (S^{1} \times S^{1}) = \mathbb{Z} \times \mathbb{Z}$

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Deformation Retractions

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