if $\pi_1(G)$ is trivial, how to prove $ \pi_1(G/H)=\pi_0(H)/\pi_0(G) $?

Question

Let $G$ be a topological group, $H$ be its normal subgroup of $G$, and $G/H$ be the quotient space induced by the natural map.(We know that $G/H$ is again a topological group)

If $\pi_1(G)$ is trivial, how to prove $ \pi_1(G/H)=\pi_0(H)/\pi_0(G) $? The author of this article use this fact directly in the page 5, but I want to know the proof of it.

http://ocw.u-tokyo.ac.jp/lecture_files/sci_03/11/notes/en/ooguri11.pdf

Plus: Can anybody show me a book of an article on the introduction to the fundamental group of topological group? It is really hard for me to find one.

Answer 1

In the article you linked there is a proof at the bottom of the page (After “To prove these relations...”), but it's obfuscated by a typo.

Whenever $F\stackrel{i}{\rightarrow} E \stackrel{p}{\rightarrow} B$ is a fibration, there is a long exact sequence in homotopy, moving downward by degree. At the end of the sequence you have $$\pi_1(G) \longrightarrow \pi_1(G/H) \stackrel{\beta}{\longrightarrow} \pi_0(H) \stackrel{i_*}{\longrightarrow} \pi_0(G) \stackrel{p_*}{\longrightarrow} \pi_0(G/H) \longrightarrow 0 $$ This is the line marked (2) in the link, but $\pi_0(H)$ is written as $\pi_0(G)$.

The assumption is that $\pi_1(G)$ is trivial, so $\beta$ is an injection. This means that $\pi_1(G/H)$ is isomorphic to the image of $\beta$. Since the sequence is exact, the image of $\beta$ is equal to the kernel of $i_*$. And the kernel of $i_*$ is the set of components of $H$ that are contained in the identity component of $G$. That is, $\pi_0(H)/\pi_0(G)$.

Answer 2

Matthew Leingang gave the answer in a comment but I'll flesh it out a little. The quotient map $q\colon G\to G/H$ is a fibration with fiber topologically isomorphic to $H$, and so we get a long exact sequence in homotopy groups $$\cdots \to \pi_{k+1}(G) \to \pi_{k+1}(G/H) \stackrel{\delta_k}{\to} \pi_k(H) \stackrel{i_*}{\to} \pi_k(G) \to \pi_k(G/H) \to\cdots.$$ Using the fact that $\pi_1(G)=0$ this gives a short exact sequence at $k=0$ which looks like $$0 \to \pi_1(G/H) \stackrel{\delta_0}{\to} \pi_0(H) \stackrel{i_*}{\to} \pi_0(G) \to \pi_0(G/H)$$ where we see that $\delta_0$ is an injection by exactness, so $\pi_1(G/H)$ is embedded as a subgroup into $\pi_0(H)$. The image of $\delta_0$ is equal to the kernel of $i_*$ by exactness, so $\ker i_* \cong \pi_1(G/H)$. But, we also know that the kernal of $i_*$ is the set of path components of $H$ which are included into the path component containing the identity element of $G$, which is precisely $\pi_0(H)/\pi_0(G)$.