Gromov-Hausdorff convergence to a circle
I am working on the book A course in metric geometry written by D. Burago, Y. Burago and S. Ivanov, and more precisely on exercice 7.5.9:
Exercice: Let $\{X_n\}$ be a sequence of compact length spaces, $X_n \underset{GH}{\to} S^1$. Prove that, for all large enough $n$, the spaces $X_n$ are not simply connected.
In fact, I solved the next exercice that implies that there exists a surjective morphism $$\pi_1(X_n) \twoheadrightarrow \pi_1(S^1) \simeq \mathbb{Z}$$ for large enough $n$. Therefore, $\pi_1(X_n)$ cannot be trivial. However, I would like to solve it using the given hint: it is not difficult to find continuous $\epsilon_n$-isometries $f_n : S^1 \to X_n$ satisfying $\epsilon_n \to 0$, and now I have to show that the loop associated to $f_n$ is not contractible in $X_n$ for large enough $n$.
Visually, it seems to be obvious, but I did not find a good approach to do this. Do you have some hint?
This question has been annoying me for quite a while. I decided to have a look at A course in metric geometry, and it turns out that your indications are somewhat misleading. The hint refers to Question 7.5.8, in which it is shown that:
There exists a sequence of continuous maps $f_n : \ X_n \to \mathbb{S}_1$ and a sequence $(\varepsilon_n)$ with $\lim_{n \to + \infty} \varepsilon_n = 0$ such that each $f_n$ is a $\varepsilon_n$-isometry,
This is convenient, because pushing forward loops is easy.
Let $(f_n)$ and $(\varepsilon_n)$ be as above. Let $n \geq 0$ and $N > 0$ be integers. Denote by $\mathbb{U}_N$ the set of $N$th roots of unity. Since $f_n (X_n)$ must be a $\varepsilon_n$-net in $\mathbb{S}_1$, we can find a sequence $(x_{n, k})_{k \in \mathbb{Z} / N \mathbb{Z}}$ such that $d(f_n(x_{n,k}), e^{2\pi i \frac{k}{N}}) \leq \varepsilon_n$ for all $k$.
Since $f_n$ is an $\varepsilon_n$-isometry, for all $k \in \mathbb{Z} / N \mathbb{Z}$,
$$d(x_{n, k}, x_{n, k+1}) \leq 2 \varepsilon_n + \frac{2\pi}{N}.$$
But $X_n$ is a length space, so we can find a curve of length at most $2 \varepsilon_n + 2\pi/N$ between those two points. By concatenating all these $N$ curves, we get a map $g_n : \ \mathbb{S}_1 \to X_n$ such that $g_n (x_{n,k}) = e^{2\pi i \frac{k}{N}}$ for all $k$. In addition, let $x$, $y$ be in $\mathbb{S}_1$. Let $e^{2\pi i \frac{k}{N}}$ and $e^{2\pi i \frac{k'}{N}}$ be one of the closest element of $\mathbb{U}_N$ for $x$ and $y$ respectively. Then:
$$|d(g_n(x), g_n(y)) - d(x,y)| \leq d(g_n(x), x_{n, k})+d(x_{n, k'}, g_n(y)) + |d(x_{n, k}, x_{n, k'}) - d(x,y)|$$
$$|d(g_n(x), g_n(y)) - d(x,y)| \leq 4 \varepsilon_n + \frac{4\pi}{N} + |d(x_{n, k}, x_{n, k'}) - d(x,y)|,$$
and:
$$|d(x_{n, k}, x_{n, k'}) - d(x,y)| \leq d(x, e^{2\pi i \frac{k}{N}}) + d( e^{2\pi i \frac{k'}{N}}, y) + | d(x_{n, k}, x_{n, k'}) - d(e^{2\pi i \frac{k}{N}},e^{2\pi i \frac{k'}{N}})|$$
$$|d(x_{n, k}, x_{n, k'}) - d(x,y)| \leq \frac{2 \pi}{N} + | d(x_{n, k}, x_{n, k'}) - d(f_n(x_{n, k}),f_n(x_{n, k'}))|$$
$$|d(x_{n, k}, x_{n, k'}) - d(x,y)| \leq \frac{2 \pi}{N} + \varepsilon_n.$$
Hence, $g_n$ is a $5 \varepsilon_n + 6\pi/N$-isometry onto its image. This is the quasi-lift we wanted to construct.
Let $\delta \in (0, 1/3)$. For all large enough $n$, I can find $N$ such that $g_n$ is a $\delta$-isometry onto its image and $f_n$ is a $\delta$-isometry, and $2\pi/n < \delta$. Then, $f_n \circ g_n$ is a $2\delta$-isometry from $\mathbb{S}_1$ to itself. Since $f_n \circ g_n$ maps $\mathbb{U}_N$ to itself,
$$\|f_n\circ g_n - id \|_{\infty} \leq 3 \delta < 1.$$
But if two maps from $\mathbb{S}_1$ to itself are at (supremum) distance strictly less than $1$, then they have the same index. Hence, $f_n\circ g_n$ has index one, and thus is not contractible.
If $(g_{n, t})_{t \in [0,1]}$ were a contraction of $g_n$ to a point in $X_n$, then $(f_n \circ g_{n, t})_{t \in [0,1]}$ would be a contraction from $f_n \circ g_n$ to a point in $\mathbb{S}_1$, which contradicts the non-triviality of the loop $f_n\circ g_n$.