# short exact sequence from fibration

Let $$G$$ be a finite group acting freely on a path connected topological space $$X$$. The covering map $$X \to X/G$$ induces a long exact sequence of homotopy groups. Since $$\pi_1(G) = 1$$ and $$\pi_0(X) = 0$$, we obtain a short exact sequence $$1 \to \pi_1(X) \to \pi_1(X/G) \to G \to 0.$$ The last map here, call it $$\partial$$, is in principle only a map between pointed sets.

Baby question: Is $$\partial$$ always a group homomorphism?

Question: If $$\partial$$ is a group homomorphism, then the above is a short exact sequence of groups. Is it always non-split? (Assuming of course that $$G, \pi_1(X) \neq 1$$. I am most interested in the case when $$\pi_1(X) = \mathbf{Z}$$.)

• If you have a closed curve at your basepoint, you lift it and the endpoint p of the lift is on the fiber of the basepoint: the image under \partial is the element of G that maps the basepoint of X to p. This is clearly a group morphism. In other words, this the map is the monodromy. Commented May 26, 2023 at 16:15
• Why do you think it should always be non-split? If you drop finiteness and only require $G$ is discrete it is easy to came out with split examples (for instance $G=\mathbb{Z}$ and $X$ an infinite cylinder so that $X/G=T^2$). Does finiteness play a a crucial role in your intuition? Commented May 26, 2023 at 17:32

For your main question, covering space theory tells us that every short exact sequence of groups $$1 \to N \to E \to G \to 1$$ occurs in this fashion. So, pick your favorite non-split example with $$G$$ finite.
Here's a few more details. Pick a path connected CW-complex $$Y$$ such that $$\pi_1(Y)$$ is isomorphic to $$E$$. Covering space theory gives us the following:
• A regular covering map $$f : X \mapsto Y$$ such that $$X$$ is path connected, and such that the induced homomorphism $$f_* : \pi_1(X) \to \pi_1(Y)\approx E$$ is injective and has image $$N$$.
• An isomorphism between the quotient group $$G \approx E/N$$ and the deck transformation group of the covering map $$f$$, hence $$Q$$ acts freely and properly discontinuously on $$S$$ with quotient space $$Y$$.
Let $$X=S^1\times S^\infty\to X/G=S^1\times\mathbb{RP}^\infty$$ where $$G=\mathbb Z/2$$. Then the short exact sequence is $$0\to\mathbb {Z}\to\mathbb {Z}\times\mathbb{Z}/2\to\mathbb Z/2\to 0$$, which is split.