TITLE: Classifying space of a group
QUESTION [1 upvotes]: My question is 
If $A$ is a subgroup of $G$ then $$BA \subseteq BG ?$$ 
where $BA$ is classifying space of $A$ and $BG$ is classifying space of $G$

REPLY [4 votes]: Classifying spaces are only well-defined up to homotopy equivalence, so a better question is whether one can construct $BA$ and $BG$ within their homotopy classes so that $BA \subset BG$. Even better, can they be constructed so that the induced map $A \approx \pi_1(BA) \to \pi_1(BG) \approx G$ is the given injection $A \hookrightarrow G$?
Here's one such construction.
Start with $BG$ with universal covering map $EG \mapsto BG$ where $EG$ is contractible. The deck transformation action of $G$ on $EG$ restricts to a deck transformation action of $A$ on $EG$, whose quotient is therefore $BA$. So there is a covering map $p : BA \mapsto BG$ such that the induced homomorphism $A \approx \pi_1(BA) \hookrightarrow \pi_1(BG) \approx G$ is the inclusion map.
Now replace $BG$ by the mapping cylinder of the covering map $p$. As with any mapping cylinder, it is still homotopy equivalent to the range of the map, which is $BG$. And now the domain of the map, which is $BA$, is included into the mapping cylinder in a way that induces the inclusion map $A \hookrightarrow G$.
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