I came across this old exam question while studying for my own exam for our topology course.
Let $f : S^1 \to S^1 $ be the map $z \mapsto z^n$. What is the induced map $$f_{*} : \pi_1 (S^1 , (1,0)) \to \pi_1 (S^1 , (1,0) ) \ \ ? $$
Could you also please explain the intuition behind induced maps, fundamental groups and how they relate to one another? I have difficulties understanding these concepts. Could you also please explain this while relating to the question at hand?
On
For $f : X \to Y$ a continuous map, the induced map $f_\ast : \pi_1(X,x) \to \pi_1(Y,y)$ is defined to be the group homomorphism mapping the homotopy class of the loop $\gamma : [0,1] \to X$ to the homotopy class of the loop $f \circ \gamma : [0,1] \to Y$.
Here, as $\pi_1(S^1,1)$ is generated by the homotopy class of $\gamma : [0,1] \to S^1, t \mapsto ???$, you just have to determine $f\circ \gamma$ to determine $f_\ast$ entirely. (I let you find the generator and its composition by $f$.)
The fundamental group of loops based at $x$ $\pi_1 (X,x)$ is intuitively the set of loops based at $x$, which can be "multiplied" by concatenation. I'm sort of lying because if two loops can be continuously deformed into one another we want to think of them as the same, so the fundamental group (as a set) is really an equivalence class of loops. The relevant equivalence relation on loops is called homotopy equivalence. It needs to be checked that this forms a group. It is an extremely important fact that $\pi_1 (S^1 , p) \cong \mathbb Z$.
As an aside, you might be wondering why we keep specifying a $x \in X$. This is because $\pi_1 (X , x)$ depends on $x$. If $X$ is not path connected, then we can potentially have $x_1, x_2 \in X$ with $\pi_1 (X , \ x_1)$ and $\pi_1 (X, \ x_2)$ not isomorphic. For example if $X$ is the disjoint union of a torus and a sphere, since choosing a point $x$ in the torus will give $\pi_1 (X, x) \equiv \mathbb Z \oplus \mathbb Z$, whereas for a point $x'$ in the sphere we have $\pi_1 (X , x')$ is trivial. For $X$ path connected, the isomorphism class of fundamental group doesn't depend on the point, but there is no canonical isomorphism between the fundamental groups with different points chosen.
The intuition behind an induced map is that if you have $f: X \to Y$ a continuous map between topological spaces, and a loop $\gamma$ based at $x \in X$, then $f(\gamma)$ is a loop based at $f(x)$. Check this, it only requires the continuity of $f$. It turns out that this is actually a group homomorphism from $\pi_1 (X,x)$ to $\pi_1 (Y, f(x))$. This is what Stefan means when he says $\pi_1$ is a functor. It maps a pointed topological space to a group, and sends continuous maps to group homomorphisms. There's a little more to say, but if you aren't familiar with functors feel free to ignore this.
Once you believe $ \ \pi_1 (S^1, (1,0)) \cong \mathbb Z$, and that $f_*$ is a group homomorphism, then by basic group theory $f_*: \ \pi_1 (S^1, (1,0)) \to \ \pi_1 (S^1, (1,0))$ must be of the form $f_* (z)= nz \ $ for some $n \in \mathbb Z$. It isn't hard to find what this $n$ must be.
Since $ \ \pi_1 (S^1, (1,0))$ is generated by the loop $\gamma(t) = e^{it}$, $0 \leq t \leq 2\pi$, that is the loop that traverses the circle once (this loop represents $1 \in \ \pi_1 (S^1, (1,0)) ) \ $ in a counterclockwise angle, then we just need to see what $f_* (\gamma)$ is. Well, $f(\gamma(t))= e^{nit}$, which traverses the circle $n$ times as $t$ goes from $0$ to $2 \pi$. So, while $\gamma(t)$ represents $1$, $f(\gamma(t))$ represents $n$.
So we see that $f_*$ is multiplication by $n$.
The best reference is Hatcher's Algebraic Topology, which is free online, and handles the fundamental group very well in Chapter 1. In there you can also find a way to think about this problem in terms of covering spaces.