Does inclusion give an injective homeomorphism $\pi[G \cap B(a, r),a] \to \pi[G, a]$? Let $G$ be a open, path connected subset of the plane and $a\in G$. Set $G' = G\cap B(a,r)$, where $B(a,r)$ is the open ball of center $a$ and radius $r$. Let $\gamma$ be a closed loop in $G'$ with $\gamma(0)=\gamma(1)=a$. If $\gamma$ is homotopic to the constant path at $a$ in $G$, is it homotopic to the constant path in $G'$?
Obviously this is not true for e.g. $G$ subset of a sphere, so it being 2-dimensional manifold is not enough.

I think I can show this is true like this, but am very curious if there are other/simpler ways to do this, or I'm just wrong. 
Lets assume that we can find a simple closed curve $C$ in $G'$ that encloses $\gamma$. Now using the Jordan–Schoenflies theorem there is a homeomorphism
$f:\mathbb{R}^2 \to \mathbb{R}^2$ such that $f(S^1) = C$. Now $D^2$ is a retract of $\mathbb{R^2}$ so there exists a continuous $r:\mathbb{R^2}\to D^2$, where $D^2$ is the closed disk of radius 1 centered at 1, with $r|D^2 = id_{D^2}$. Let $F$ be a homotopy from $\gamma$ to the constant path at $a$ in $G$. Then 
$$ h = f \circ r \circ f^{-1} \circ F$$ is a homotopy in $G'$.
I dont like this assumption about existence of the closed curve $C$ and I don't know how to show it.
 A: Here is a proof that $\gamma$ is homotopically trivial in $G'$. It uses the same elements as your proof, i.e. Jordan-Schoenflies and a winding number argument.
We may assume $B(a,r) \not\subset G$ else we are done.
Let $h : D^2 \to G$ be a continuous map which extends $\gamma$. 
By perturbing $h$, we may assume that it is in general position with respect to the circle $S_r = \partial \bar B(a,r)$, and so $h^{-1}(S_r)$ is a finite union of pairwise disjoint circles in the interior of $D^2$. Let the number of these circles equal $N$. I'll induct on $N$. If $N=0$, we're done.
Assuming $N > 0$, let $b_1,...,b_N$ be the components of $h^{-1}(S_r)$. These are boundaries of discs I'll denote $d_1,...,d_N$ (here the Jordan-Schoenflies theorem must be used).
Let $b_i$ be a component of $h^{-1}(S_r)$ which is innermost, meaning that $d_i$ contains no other $b_j$. It follows that either $h(d_i)$ is on the inside meaning $h(d_i) \subset \overline B(a,r)$, or it is on the outside meaning $h(d_i) \subset \mathbb{R}^2 - B(a,r)$.
Now we argue in two cases, depending on the degree $D$ of the map $h \, | \, b_i : b_i \to S_r$. 
If $D=0$ then $h \, | \, b_i$ is homotopic to a constant within $S_r$. We can therefore redefine the map $h$ on $d_i$ to take values in $S_r$; let that redefined map be denoted $h'$. We can then perturb $h'$, moving $h'(d_i)$ entirely off of $S_r$, either pushing it to the inside of $S_r$ or to the outside of $S_r$ depending on whether $h(d_i)$ was on the outside or on the inside to begin with; let that perturbed map be $h''$. This map $h'' : D \to G$ is still a continuous extension of $\gamma$, and it is still in general position with respect to $S_r$, but the number of components of $(h'')^{-1}(S_r)$ is $N-1$, completing the induction.
If $D \ne 0$, I'll argue to one of a few contradictions. The map $h | b_i : b_i \to S_r$ must be surjective, and in particular it follows that $S_r \subset G$. If $h(d_i)$ is on the outside of $\bar B(a,r)$ then $h | b_i$ is homotopically nontrivial on the outside, which is absurd; this holds because the outside is an annulus and the inclusion of $S_r$ to this annulus is a homotopy equivalence (this is the step that fails in the 2-sphere). Thus $h(d_i)$ is on the inside of $\bar B(a,r)$. But now we can use a winding number argument, with the conclusion the map $h | d_i : d_i \to \bar B(a,r)$ must be surjective. It follows that $\bar B(a,r) \subset G$, contrary to our assumption.
