I don't understand what happens in the induced homomorphism of a universal covering map.
A covering map $p:\tilde X \to X$ is universal if $\tilde X$ is simply connected, so the fundamental group $\pi_1(\tilde X, \tilde x_0)$ = $\{ 1 \}$.
But then the induced homomorphism $p_*$ is mapping $\pi_1(\tilde X, \tilde x_0) = \{ 1 \}$ to the fundamental group $\pi_1(X, x_0)$ of $X$, so surely $\pi_1(X, x_0) = \{ 1 \}$ also?
But then we have $\exp: \mathbb{R} \to S^1$ is a universal cover of $S^1$ which induces a homomorphism $\exp_* : \{1\} \to \mathbb{Z}$ which I thought was impossible!
What am I doing wrong?
Well... Let $p$ be a covering. Unless $p$ is a homeomorphism, $p$ doesn't induce a surjective homomorphism between the fundamental groups. Why is it so? Because all coverings induce injections between the fundamental groups. So, if $p$ is a covering and induces a surjection between fundamental groups, you have that this $p$ induces isomorphisms between fundamental groups. In this case, you would have that $p$ is a homeomorphism.
When you have a universal covering, you have nice results. First of all, $Aut(p)\cong \pi _ 1(B,b) $.
And, about the universal covering of the circumference, you may see my answer at Covering space and Fundamental group