Let $\delta$ in $(\mathbb{R}^2-\alpha)$ where $\alpha$ is a path joining $0$ to $\infty$ which is injective, i.e., it is a simple path, it does not intersect itself. Show that $\delta$ is homotopic to a constant path in $(\mathbb{R}^2-{0})$. I am assuming the path are defined on the interval $[0,1]$.
Here, homotopic means that there is a continuous application from $\delta$ to the constant path, with fixed extremities, i.e. I should be able to crunch $\delta$ to a point without moving one of its point which we will assume to be $\delta(0)=\delta(1)$ (since $\delta$ is closed). Also, obviously, the crunching should not involve an intermediary path crossing $0$.
This means, assuming the paths are parametrized on the interval $[0,1]$, that I should be able to find a continuous application $H$ from $[0,1]^2$ to $\mathbb{R}^2-0$ such that $H(0,t)=\delta(t), H(1,t)=\delta(0), H(s,0)=\delta(0)=H(s,1)$ (this last double equality encodes fixed extremities, the $H(s,t)$ are intermediary paths which should continuously go from $\delta$ to the constant path)
A hint is given, telling me to consider $\delta(t)-\alpha(s)$, but I don't know what to do from here, since it does not even fix extremities and goes to infinity. It does not intersect $0$ though, which is a good thing at least. What can I pick for $H$ that will work for any path $\delta$? If its interior is convex then this is easy, but otherwise this can become complicated...
This doesn't directly give a homotopy $H$, but the argument is at least elementary. Consider, as in the hint, the family of loops $$\gamma_s: S^1 \to \mathbb{R}^2 \setminus \{0\}, \quad t \mapsto \delta(t) - \alpha(s)$$ for $s \in [0, \infty)$.
Since $\alpha(s) \to \infty$ and the image of $\delta$ is bounded, there exists $S \geq 0$ such that $\lvert \alpha(S) \rvert > \lvert \delta(t) \rvert$ for all $t$. I claim that the degree of the map $\gamma_S: S^1 \to \mathbb{R}^2 \setminus \{0\} \simeq S^1$ is zero. If not, then for some $t$, $\delta(t) - \alpha(S)$ must be perpendicular to $\alpha(S)$, i.e., $0 = \langle \delta(t) - \alpha(S), \alpha(S) \rangle = \langle \delta(t), \alpha(S) \rangle - \lvert \alpha(S) \rvert^2$. But by Cauchy-Schwarz, we have $\lvert \langle \delta(t), \alpha(S) \rangle \rvert \leq \lvert \delta(t) \rvert \cdot \lvert \alpha(S) \rvert < \lvert \alpha(S) \rvert^2$ for all $t$ by our choice of $S$. So indeed $\deg \gamma_S = 0$.
Now, the function $[0,\infty) \to \mathbb{Z}$, $s \mapsto \deg \gamma_s$ is locally constant. Since $[0,\infty)$ is connected, we deduce that in particular $\deg \gamma_0 = \deg \gamma_S = 0$. Since $\gamma_0 = \delta: S^1 \to \mathbb{R}^2 \setminus \{0\}$, this means that there is a homotopy from $\delta$ to a constant path in $\mathbb{R}^2 \setminus \{0\}$.