Neighbourhood retract of a loop Suppos $f:S^1\to M$ is a smooth map and $M$ is a smooth manifold. Does there exist a neighborhood $\mathcal N$ of the image $f(S^1)$ which retract (or deformation retract) to $f(S^1)$? If so, is there a simple proof? (I actually only need to find a neighbourhood such that $H_2(\mathcal N)=0$).
Thoughts: The set where $f$ is an immersion is an open set, so it is a disjoint union of intervals. On the complement of these intervals, we have $\nabla f=0$, i.e. the curve is "stationary". Hence, $f(S^1)$ is mostly an immersed curve, with possibly infinitely many cusps. And I am not sure how to proceed.
More thoughts: We can choose an atlas of $M$ such that $f$ can be $C^0$-approximated by a PL function $\tilde f:S^1\to M$ on some fixed charts. If $\dim M\ge 3$, we can perturb $\tilde f$ so that $S^1$ is properly embedded  in $M$. Then we can pick a regular neighborhood $\mathcal N$ of $\tilde f(S^1)$. In this case, $H_2(\mathcal N)=0$. In the case where $\dim M=2$, we can embed $M$ to $\mathbb R^k$, and the perturb curve $\tilde f:S^1\to \mathbb R^k$ will live as an embedded curve in $\mathbb R^k$. Choose a regular neighborhood of it, and since $\tilde f$ is $C^0$ close to $f$, this neighborhood contains $f(S^1)$. I think the intersection of this neighborhood with $M$ works.
I still have no idea how to show retract/deformation retract.
 A: *

*You are asking several different questions, I will answer the one which appears in the title of your post:

There are $C^\infty$-loops $f: S^1\to {\mathbb R}^n$ (for all $n\ge 2$) such that the image $f(S^1)$ is not an NR (neighborhood retract). The simplest examples I know are given by Hawaiian Earrings.
In order to ensure the existence of a $C^\infty$ parameterization of this union of circles, take the union of circles $C_k$ whose diameters are $2^{-k}$, $k\in {\mathbb N}$.

Let $H$ denote the Hawaiian Earrings. Then $\pi_1(H)$ is uncountable. If $U$ is a (connected) neighborhood of $H$ in ${\mathbb R}^n$ which admits a retraction to $H$, then the inclusion map $\pi_1(H)\to \pi_1(U)$ is injective. However, connected manifolds have countable fundamental groups. A contradiction.


*As for the existence of a neighborhood with $H_2=0$, I am not sure. I suspect, you need much less, namely, you need a system of nested neighborhoods $U_k, k\in {\mathbb N}$ of the image $K=f(S^1)\subset M$ such that:


*$\bigcap_k U_k= K$.


*For each $k$ the inclusion map $U_{k+1}\to U_k$ induces zero map  $H_2(U_{k+1})\to H_2(U_k)$.
This is indeed true and follows from several observations:
(a) Hausdorff dimension of $K$ is $\le 1$ (it suffices to know that $f$ is Lipschitz, you do not need smoothness).
(b) Hence, covering dimension of $K$ is $\le 1$.
(c) Hence, $\check{H}_2(K)=0$ (I am using Chech homology here).
(d) Hence, there exists a neighborhood system as above. You can take, for instance, $U_k$ equal to the open $1/k$-  neighborhood of $K$ in $M$ with respect to some fixed Riemannian metric on $M$.


*Your attempted proof of existence of a neighborhood with $H_2=0$ fails because you are not using the assumption that $f$ is smooth, all you are using is continuity of $f$. However, there are continuous surjective maps $S^1\to S^2$.

