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I am not sure how to proceed with this problem:

Let $X$ be a path-connected and locally path-connected space. Suppose $p:X\to X$ is a non-injective covering map. Prove that $\pi _1(X) $ is infinite for any choice of base point $a\in X$.

Pick $a\in X$ arbitrarily. I know that since $X$ is path connected, $| p^{-1}(a)| \leq |\pi _1(X,a) | $. Then since $p$ is a non-injective covering map, $|p^{-1}(a)|>1 $. (In fact, all fibers have the same cardinality.)

Is there some way to show $|\pi _1 (X,a)| $ is infinite by using deck transformations?

Or, is it possible to show that if a space covers itself, then its fundamental group is either trivial or infinite?

Any suggestions would be helpful.

RobPratt
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J-Q-McNeedHelpson
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2 Answers2

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You know that $| p^{-1}(a)| \leq |\pi _1(X,a) | $. Thus, if the fiber $p^{-1}(a)$ is infinite, we are done. So let us consider the case of a finite fiber with $n$ elements. We can compose $p$ with itself arbitarily many times to get surjections $p^m : X \to X$. It is well-known that they are coverings; see Question on covering spaces . The fiber of $p^m$ has $n^m$ elements. It is well-known that $p^m_*(\pi_1(X,a))$ is a subgroup of $\pi_1(X,a)$ of index $n^m$. Since $m$ is arbitrarily large, we conclude that $\pi_1(X,a)$ must be infinite.

Paul Frost
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Let $F$ be the fibre of the fibration. The exact sequence defined by the fibration gives $1=\pi_1(F)\rightarrow\pi_1(X)\rightarrow\pi_1(X)\rightarrow\pi_0(F)\rightarrow\pi_0(X)=1$. It result that $\pi_1(X)$ has a subgroup isomorphic to itself which is not itself since $\pi_0(F)$ is not trivial. This implies that the cardinality of $\pi_1(X)$ is infinite, since a subset of a finite set distinct of itself has a strictly inferior cardinality.

Tsemo Aristide
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