What is a pointed covering (topology)?

Question

So our prof has defined a covering map/space as in wiki. After that, in a lemma, he just used the term "pointed covering map" without defining it. I'm not sure if this is the correct english term, I had to translate it from German.

This is the context:

Lemma: Let $p: (\bar X , \bar x ) \rightarrow (X,x)$ be a pointed covering map. Then the homomorphism $\pi_1(p, \bar x ): \pi_1(\bar X , \bar x ) \rightarrow \pi_1(X,x)$ is injective.

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More context. I have an exercise to solve:

Let $p: \bar X \rightarrow X$ be a covering of path connected spaces. Let $\pi_1(X)$ be a finitely generated free group. Proove that $\pi_1(\bar X )$ is also finitely generated free group.

I am confused if I am allowed to apply the lemma here or not, because we are not given a pointed covering map, but only a "normal" covering map, and as said, I'm not sure what the actual definition of a pointed covering map is.

However, suppose we are allowed to apply it, e.g. by just chosing arbitrary basepoints $\bar x \in \bar X $ and $p(x)\in X$. Then since both spaces are path connected, the fundamental group is independent of our choice. But then this exercise is trivial: Apply the lemma, so $\pi_1(\bar X )$ can be identified as a subgroup of $\pi_1(X)$ and is therefore finitely generated free group.

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Can someone help me understand this?

Answer 1

A pointed covering map $p: (\bar X , \bar x ) \rightarrow (X,x)$ is in fact nothing else than a pointed map $p: (\bar X , \bar x ) \rightarrow (X,x)$ such that the "free" map $p : \bar X \to X$ is a covering map.

That is, if we have a covering map $p : \bar X \to X$, we can choose any basepoint $x \in X$ and any $\bar x \in p^{-1}(x)$ to obtain a pointed covering map $p: (\bar X , \bar x ) \rightarrow (X,x)$.

Basepoints are needed if we want to deal with fundamental gropus and induced homomorphisms.

You are right, if $X$ and $\bar X$ are path-connected, then the fundamental groups are independent (up to isomporphism) of the choice of basepoints. In that sense your exercise is trivial: By the lemma you know $\pi_1(\bar X)$ is isomorphic to a subgroup of $\pi_1(X)$. If you also know the purely group theoretic result that subgroups of free groups are free, you are done.

However, note that we can also use the theory of covering projections to prove that subgroups of free groups are free. See for example Chapter 3 Section 8 of

Spanier, Edwin H. Algebraic topology. Springer Science & Business Media, 1989.