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!
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.