Induced homomorphisms in a universal covering map

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?

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The morphism $p_*$ needn't be surjective, so you can't conclude that $\pi_1(X, x_0) = \{ 1 \}$ . –  Georges Elencwajg Apr 12 '13 at 12:44
Just because $X\to Y$ is surjective, doesn't mean $\pi(X)\to\pi(Y)$ is surjective. –  Thomas Andrews Apr 12 '13 at 13:12

1 Answer

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

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