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Apparently, the definition of homotopy formalizes the idea of continuous transformation between two things.

(*) Let's take this motivating example: I have $ S,S' \subset \mathbb{R}^3 $ two surfaces in space. If I wanted to define a continuous transformation between them I would request for a continuous function $\Gamma:S\times [0,1] \rightarrow \mathbb{R}^3$ to exist such that $\Gamma(\cdot,0)=id|_S $ and $Im(\Gamma(\cdot,1))=S' $.

Instead, the definition in use is not between spaces ($S$ and $S'$) but between functions. I think my definition (*) would be a special case of the definition in use. In fact, by posing $f:=\Gamma(\cdot,0)=id|_S$ and $f':=\Gamma(\cdot,1)=id|_{S'}$ I have that $f\simeq f'$.

Can someone clarify what's the limitations (or what's wrong with) my definition (*)?

Eric Wofsey
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rod
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    I think my doubt arises from the fact that I don't see the difference between "continuous transformation of $f$ into $f'$" and "continuous transformation of $Im(f)$ and $Im(f')$". – rod Feb 01 '21 at 20:53
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    You can't have $\Gamma(\cdot,0)=id_{|S}$ unless $S \subset S'$. – Aphelli Feb 01 '21 at 21:10
  • Thanks, $\Gamma$ was meant to go to the ambient space. I'll correct it. – rod Feb 01 '21 at 21:13
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    There's notions called isotopy and ambient isotopy which are close to what you seem to be looking for! – Jeroen van der Meer Feb 01 '21 at 21:21
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    @roddik You can visualize it by the continuous transformation of the image of $f$ to the image of $f'$. – Berci Feb 01 '21 at 21:30
  • @Berci yes, that's the point. Why don't we use my (*) definition then? – rod Feb 01 '21 at 21:31
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    Here’s a problem with ambient spaces: there may be a continuous surjective map $f:S \rightarrow S’$ (if $S$ is a torus and $S’$ is a sphere, for instance). Define $\Gamma(x,t)=(1-2t)x$ if $t \leq 1/2$ and $\Gamma(x,t)=(2t-1)f(x)$ if $t \geq 1/2$. Then you deform continuously $S$ into $S’$ – even though their topological properties are very different (in my example, one has a “hole”, the other hasn’t). – Aphelli Feb 01 '21 at 21:54
  • @Mandlack The problem there is apparently that at $t=1/2$ the transformation collapses into a point, and then restarts widening. Are you saying that my definition is allowing this, while that (correct) of homotopy is not? – rod Feb 01 '21 at 22:11
  • With your definition any two surfaces in $R^3$ will be equivalent (in your sense), because $R^3$ is contractible. What would this be good for? – Moishe Kohan Feb 02 '21 at 00:00
  • Well of course the codomain of the functions $f,,f'$ would differ (say, a torus) - and in that case $S,S'$ would be two lines/loops on the torus – rod Feb 03 '21 at 11:18

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I think this is a very nice question, and your definition (*) is very close to the definition of an isotopy. You want to also require that $\Gamma(S,t)$ is an embedding for all $t$ so that you can't collapse everything to a point (as was pointed out in the comments). Otherwise, any two maps would be equivalent.

I think it's useful as well to compare isotopy to the notion of homotopy equivalence. As an example the circle is homotopy equivalent to itself via the identity map, but it has many distinct (up to isotopy) embeddings in $\mathbb{R}^3$. Specifically, the unknot and trefoil can be shown to be non-isotopic using various knot invariants.

Strongly Negative Amphicheiral
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  • Thanks for your answer. So is my definition + the request of $\Gamma$ being an embedding at each time equivalent to the definition of an isotopy? – rod Feb 01 '21 at 22:27
  • Also, why the standard definition doesn't allow the "collapse"? I don't see the difference in that sense. – rod Feb 01 '21 at 22:29
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    I think your definition plus the request of being an embedding at each time is exactly the definition of an isotopy. For the second question, I'm not sure which standard definition you're referring to. For homotopy equivalence, the maps are not embeddings anywhere, so there's no ambient space to 'shrink to a point' in. – Strongly Negative Amphicheiral Feb 01 '21 at 22:45
  • Well I was taught that $f,f':X\rightarrow Y$ are homotopic if $F:X\times[0,1]\rightarrow Y$ (...) $F|_0=f$ and $F|_1=f'$. In my example the sets continuously mouved would be $Im(f)$ and $Im(f')$ so the ambient space would be $Y$. Does this make sense? – rod Feb 01 '21 at 22:51
  • You are correct that those two maps are homotopic; any two maps to a contractible space (e.g. R^3) will be homotopic, exactly by contracting them both to a point as Mindlack pointed out above. In that sense homotopy really should be thought of as a transformation of maps, not spaces. – Strongly Negative Amphicheiral Feb 01 '21 at 23:24