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Finding infinitely many nonhomotopic retractions r: $S^1 \vee S^1 \rightarrow S^1$

So, in theory would it be possible to find infinitely many nonhomotopic retractions with the domain just being $S^1$ rather than $S^1 \vee S^1$?

Like $e^{ix} \rightarrow e^{nix}$ would be a unique retraction for each n? Then we could extend this to the wedge by having it act the same way for the other one? My logic is probably flawed here. Can anyone offer some general insight into the situation please? Thanks!

  • I think you are on the right track! if you can prove these maps are not homotopic to each other, this method should work. – Rylee Lyman Feb 04 '19 at 02:00
  • Check my answer to the similar post https://math.stackexchange.com/questions/2987992/nonhomotopic-retractions-s1-vee-s1-to-s1-without-seifert-van-kampen/2988435#2988435 – Tyrone Feb 04 '19 at 10:21

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A retraction is required to be the identity on its image so you will not be just copying one map onto both circles. What you can do though, is set the map equal to the identity on one circle and map the other using the map you describe.

If you must be rigorous with showing that these maps are not homotopic, a direct approach might not be the cleanest (assuming you want them not homotopic as maps $S^1 \vee S^1 \rightarrow S^1 \vee S^1$).

Here is how I would proceed:

We know that the wedge product is the coproduct of basepointed spaces and continuous maps between them. It can be checked that it is also the coproduct of basepointed spaces with homotopy classes of maps between them. This means that $\langle S^1 \vee S^1 , S^1 \vee S^1 \rangle $ decomposes as $\langle S^1 , S^1 \vee S^1 \rangle \oplus \langle S^1 , S^1 \vee S^1 \rangle $ . We know from van Kampen's theorem how $\langle S^1 , S^1 \vee S^1 \rangle $ decomposes. Collecting this we have the homotopy classes of maps are $(\mathbb{Z}* \mathbb{Z}) \oplus (\mathbb{Z}*\mathbb{Z})$, and we know what the representatives are for the $\mathbb{Z}$'s. Now we check what each of the retractions correspond to: $(n',1')$ where the $'$ denotes an element of the fundamental group of the circle we are retracting to. So for different $n$ these are not homotopic.

Connor Malin
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  • I'm not up to date on terminology like "coproducts" as of right now. I was thinking about doing what you said where you set the map equal to the identity on one of the circles, but I was a bit worried because then would the map be well defined on the point where the circles touch? (i mean, spot where the wedge product is formed). Furthermore, I assumed that we were supposed to show that they were nonhomotopic as maps onto $S^1$ –  Feb 04 '19 at 13:27
  • So coproduct just means that maps out of it correspond to products of the maps out of the underlying objects. The reason I said this is because if we were talking about our retraction having codomain $S^1 \vee S^1$ we need to make sure that this extra circle, even though it is not in the image, does not affect the homotopy type of our retractions. Van Kampen tells us how to do this when the domain is just a circle, but not when the domain is a wedge of circles. That is why I mentioned the coproduct. – Connor Malin Feb 04 '19 at 15:46
  • If your codomain is just $S^1$ there is no need to mention coproducts. You can instead consider restrictions of homotopies and apply what you know about the fundamental group of the circle. – Connor Malin Feb 04 '19 at 15:48
  • This distinction is important because of the codomain is just $S^1$ you are allowing yourself less room to work with and you might get different homotopy classes. However, it actually is true that if $f,g: X -> A$ are retractions they are homotopic iff they are homotopic as maps into $X$. This is very easy to prove. – Connor Malin Feb 04 '19 at 15:49