Neighborhood deformation retract for the image of a curve

differential-geometry differential-topology

No in general. Such an open set might not exist even if $\gamma$ is immersed. As an example, let $\gamma : [-1, 1] \to \mathbb R^2$ satisfies

$$\gamma (t) = \begin{cases} \left( t, \sin (1/t) e^{-1/t}\right), & t\in (-1/4, 1/4) \\ (t-1/2, 0), & t\in (1/3, 2/3) \end{cases}$$ and otherwise $\gamma(t)$ is outside a fixed ball $B_1$ centered at the origin with $\gamma'(t) \neq 0$.

Let $V$ be any open set containing the image $\Gamma$ of $\gamma$. If $r: V\to \Gamma$ is a retraction, let $B$ be an open ball centered at the origin so that $B \subset V \cap B_1$. Then $r|_B : B\to B\cap \Gamma$ is a retraction, which is not possible since $B\cap \Gamma$ has nontrivial fundamental group (indeed, it still has infinitely many holes).

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