TITLE: Prove that an annulus is not simply connected?
QUESTION [6 upvotes]: I don't have complex analysis at my beck and call, and I only have a low level of knowledge in topology, but I need to prove that this metric space (for any real $r$ and $R$ with $r < R$)$$ X = \{ (x, y) \in \mathbb{R}^2 \ | \ r \leq x^2 + y^2 \leq R \}$$
with the Manhattan metric $d((x_1, y_1), (x_2, y_2)) = |x_1-x_2| + |y_1-y_2|$ is not simply connected.
I've already prooven that it's path connected, and now I need to show there are some points $P$ and $Q$ with two paths between them such that one cannot be continuously 'morphed' into the other.
$ \ $
What I have so far is as follows:
I take $P = (0, r)$ and $Q = (0, -r)$, with $f_0$ being a path from $P$ to $Q$ going clockwise around the circle radius $r$ and $f_1$ being much the same but going counterclockwise.
Now I assume there is a function $g : [0, 1]^2 \to X$ such that:

$g(s, 0) = f_0 (s)$
$g(s, 1) = f_1 (s)$
$g(0, t) = P$
$g(1, t) = Q$

To get the final result, I need to show that this function cannot be continuous, but for the life of me I cannot.
For some context, these are the topics which have been visited during the course, roughly in order of recentness.

Multiple connectedness
Simple connectedness
Pathwise connectedness
Interior points
Boundary points
Open sets
Compactness
Complete metric spaces
Bounded metric spaces
Totally bounded metric spaces
Closed sets
Closure of a metric space
Limit points
Cauchy sequences
Convergence
Continuity
Metric space quivalence
Metric equivalence

Edit: I might have an argument that works, though it's far from rigourous. We can shrink $R$ to be as close to $r$ as we want, so we can essentially constrain the annulus down to a circle and thus force any path from the left side of the circle to the right side to go through $P$ or $Q$.
So holding $s \in (0, 1)$ constant and varying $t$ must produce a path through $P$ or $Q$ for any $s$. If for some $s$ it passes through $P$ and for some other $s$ it passes through $Q$, then there must be $s_0$ such that $\forall \epsilon >0  \ \exists \delta \leq \epsilon$ st. $s_0$ produces a path through $P$ and $s_0 + \delta$ produces a path through $Q$.
Now we can consider the path $g(s, \frac{1}{2})$, and note that it must have a discontinuity at $s_0$.
Now consider the case where all $s$ produce paths through only one of $P$ or $Q$. WOLOG: $P$. Now, at $t = \frac{1}{2}$, $s$ arbitrarily close to $1$ are mapped away from $Q$, but $1$ is always mapped to $Q$ by definition, so $g(s, \frac{1}{2})$ has a discontinuity at $s=1$. Therefore $g(s, t)$ is not continuous.
This argument is definitely iffy to me, if no one has their own argument (using sufficiently low level concepts), then criticism on the above would be appreciated.

REPLY [2 votes]: Hint: Compute the path integral
$$
\int\frac{x\,\mathrm{d}y-y\,\mathrm{d}x}{x^2+y^2}\tag{1}
$$
along the two paths from $(\sqrt{rR},0)$ to $(-\sqrt{rR},0)$ parametrized by $(\sqrt{rR}\cos(\theta),\sqrt{rR}\sin(\theta))$ for $\theta\in[0,\pi]$ and $\theta\in[0,-\pi]$.
Use Green's Theorem to show that, on any closed contour which is the difference of two neighboring paths inside the annulus, the integral in $(1)$ is $0$. Thus, if you can continuously deform one path to another inside the annulus, the change of the integral along the paths would be $0$.

A simple corollary of Green's Theorem is that if a region is simply connected, and
$$
\frac{\partial}{\partial x}F(x,y)=\frac{\partial}{\partial y}G(x,y)\tag{2}
$$
at every point in that region, then, over any closed path $\gamma$ in that region,
$$
\int_\gamma F(x,y)\,\mathrm{d}y+G(x,y)\,\mathrm{d}x=0\tag{3}
$$