The fundamental group of an inverse limit of an inverse system of topological spaces

Question

If I have an inverse system of topological spaces, which fundamental group is $\mathbb{Z}$, would the fundamental group of its inverse limit be $\mathbb{Z}$? And if not, under what conditions is it true? I'm very new to this topic, so I would be grateful, if you give me an explanation in a basic way and point out, what I should read to understand the topic.

Answer 1

A standard counterexample is given by dyadic solenoid, which has embeddings into $\mathbb R^3$ as shown in that link, but which also has a simple construction as the inverse limit of the following sequence: $$\cdots \xrightarrow{f_3} S^1 \xrightarrow{f_2} S^1 \xrightarrow{f_1} S^1 $$ where $f_i(z)=z^2$ for all $i$; here I am using a complex parameter $z=x+iy$ for points of $S^1$.

The dyadic solenoid is not path connected, in fact it has uncountable many path components. Every path component admits a continuous bijection from $\mathbb R$ that satisfies path lifting and homotopy lifting, and hence every path component is simply connected.

If you read more in the wikipedia link provided you will see that there are many other solenoids to be constructed in analogous manner, just by varying the maps $f$. In fact, you can replace each individual $f_i$ with some power function $z \mapsto z^{n_i}$, $|n_i| \ge 2$, where $n_i$ varies arbitrarily as a function of $i$. All of these have simply connected path components, by the exact same construction as for the dyadic solenoid. You can even allow some of the $n_i$ to be $\pm 1$, as long as infinitely many of them are not equal to $\pm 1$, and you'll get the same conclusions. But, if all but finitely many of the $n_i$ are equal to $\pm 1$ then the inverse limit is homeomorphic to the circle and hence has fundamental group $\mathbb Z$.