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Suppose we have a space $X$ that is path connected and locally path connected, and let $f: X \rightarrow Y$ be continuous. How do we show that the following statements are equivalent?

(A) $f$ lifts to the universal cover $p: \tilde{S^1} \rightarrow S^1$, i.e. there is a continuous map $\tilde{f}: X \rightarrow \tilde{S^1}$.

(B) $f$ is nulhomotopic.

(C) $f$ induces the trivial map on fundamental groups.

I think (A) implies (C), but I'm not so sure how. Other than that, I'm not sure exactly where to start. I would appreciate some assistance here.

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up vote 1 down vote accepted

To get you started:

First, $Y$ should be $S^1$ and $\tilde{f}$ should map into $\tilde{S^1}$.

Second, you're correct that A almost immediately implies C: first recall that the universal cover of $S^1$ is $\mathbb{R}$, so if $f$ lifts to $\tilde{f}$, then $ f=p\circ \tilde{f}$ and $\tilde{f}:X\to\mathbb R$. Since $\mathbb{R}$ is contractible, $\tilde{f}$ is the zero map on $\pi_1$. Hence $f$ itself is trivial on $\pi_1$. Essentially, this is an application of one direction of the lifting criterion for covering spaces (see page 61 of Hatcher).

The other direction of the lifting criterion gets you from C to A.


Two homotopic maps will induce the same map on $\pi_1$. Hence, B implies C. A implies B because $\mathbb{R}$ contractible means that $\mathbb{R}$ deformation retracts to any point. You can then post compose your homotopy with $p$ to get a nullhomotopy of $f$. This gives you B.

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What about B? How does that come into play? – Libertron Jan 15 '13 at 17:17
I've added hints to equate B. Cheers! – BGW Jan 15 '13 at 19:33

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