Prove that an annulus is not simply connected? 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.
 A: Assume you have a continuous $g$ as stated by you.
Show that, for each $t \in [0,1]$, there exists a continuous function $\theta_{t} : [0,1]\rightarrow\mathbb{R}$ such that $\theta_{t}(0)=-\pi/2$ and such that $g(s,t)=|g(s,t)|(\cos\theta_{t}(s),\sin\theta_{t}(s))$. (Connectedness of $[0,1]$ can be helpful.) Then show that $\theta_{t}$ is unique by connectedness of $[0,1]$. Use known, simple contours for $f_{0}$ and $f_{1}$ in order to arrange for $\theta_{0}(1)=\pi/2$ and $\theta_{1}(1)=-3\pi/2$. Show that, for fixed $s \in [0,1]$, $\theta_{t}(s)$ is a continuous function of $t$ (because of homotopy,) and note that $\theta_{t}(1)$ has possible values $\pi/2\pm 2n\pi$ for $n=0,1,2,3,\ldots$. Use connectedness of $[0,1]$ to reach a contradiction.
A: 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,
$$
\int_\gamma F(x,y)\,\mathrm{d}y+G(x,y)\,\mathrm{d}x=0\tag{3}
$$
