Let $G$ be a connected Lie group, and let $H\subset G$ be a (closed) Lie subgroup, not necessarily connected. Set $X=G/H$. The fibration $j\colon G\to X$ with fiber $H$ induces an exact sequence $$ \pi_1(H)\overset{i_*}{\longrightarrow} \pi_1(G)\overset{j_*}{\longrightarrow}\pi_1(X) \overset{\delta}{\longrightarrow}\pi_0(H)\to 1. $$ We know from the general theory that the maps $i_*$ and $j_*$ are homomorphisms relating the groups $\pi_1(H)$, $\pi_1(G)$ and $\pi_1(X)$. However, in our case $\pi_0(H)$ is also a group! I need a reference for the fact (certainly well-known) that $\delta$ is a homomorphism. (I can prove it, but I would like to have a reference.) Actually, I need only the case when $H$ is a discrete subgroup in $G$; then $\pi_0(H)=H$.
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asked Apr 27, 2015 at 12:20
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4$\begingroup$ You can continue the fibre sequence as $H\xrightarrow{i} G\xrightarrow{j} X\xrightarrow{k} BH\xrightarrow{l} BG$, and then $\delta=\pm k_*$. $\endgroup$– Neil StricklandCommented Apr 27, 2015 at 13:58
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$\begingroup$ You're not even using $G$ a group, just the map $\pi_1(X/H) \to H$ from the usual theory of covering spaces (here $X \to X/H$). $\endgroup$– Allen KnutsonCommented Apr 27, 2015 at 21:41
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