Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am primarily a student of physics and am trying to self-learn some algebraic topology. I am having some difficulty understanding the differences between the constructions of

$(X,A)$ (Pair of spaces), $X/A$ (Quotient space), $G/H$ (Quotient group of topological groups), $G/H$ (Orbit space where H is viewed as acting on $G$ say by left multiplication)

My questions are as follows:

  1. If $G$ is a topological group and $H$ is a (normal) subgroup then is the quotient group $G/H$ (topologically) the same as $G/H$ viewed as a quotient topological space? If not is there a condition on the topologies or spaces in which they coincide? How does the orbit space $G/H$ differ from these two notions?

  2. I think I always took for granted that $(X,A)$ was the same as $X/A$ (quotient space) due to excision in homology but now that I am learning some homotopy theory I am not so sure. Is $(X,A)$ ever the same as $X/A$?

  3. Under what conditions is $\pi_n(X,A) \cong \pi_n(X/A)$

share|improve this question
    
For 1, it would depend on the action of H, usually there is more than one way to define an action, and its left actions by multiplication that will make the space of orbits and the quotient group the same (think of orbits as cosets). –  AnonymousCoward Oct 15 '11 at 1:11
add comment

1 Answer 1

Let's tackle number $1$ first. $G/H$ viewed as a quotient space in the topological sense is very different from the quotient group. For example, consider $G=\mathbb R^2$ as an abelian group under addition. Let $H=\mathbb R\times\{0\}$. Then $G/H$ as a topological space is just $\mathbb R^2$ with the $x$-axis shrunk to a point. As a group $G/H$ is homeomorphic $\mathbb R$. So the topological quotient is almost everywhere $2$-dimensional, whereas the group quotient is $1$-dimensional.

For number $2$ you are probably thinking that $H_n(X,A)\cong H_n(X/A)$ in many circumstances, so that roughly "$(X,A)=X/A$". The problem here is that $H_n(X,A)$ is a notational convention for relative homology based on chains $C_n(X,A)=C_n(X)/C_n(A)$. We are not thinking of $(X,A)$ as being itself a space. So really the question doesn't make semantic sense.

As for number $3$, usually one assumes that $A$ is a closed subspace with an open neighborhood that deformation retracts onto $A$. Then your statement is true.

share|improve this answer
    
In 3 it is also worth noting that if (X, A) is a relative CW complex, then your condition automatically holds; in fact, there exists such a neighborhood that A is its strong deformation retract. –  Alexei Averchenko Oct 15 '11 at 2:27
    
@Grumpy: Hello. I don't believe that your answer for number 3 is correct, since $\pi_n(X,A)$ is a statement about the htpy fiber while $\pi_n(X/A)$ is a statement about the htpy cofiber (when as you say, $A \to X$ is a cofibration). The htpy fiber and cofiber do not agree in general, and excision tells us the connectivity of the natural map $F i \to \Omega C i$ between the htpy fiber and loops on the htpy cofiber. I believe a counter example is the following pair $(D^2,S^1)$, where $\pi_n(D^2,S^1) = 0$ for $n \geq 3$ by the l.e.s., while $\pi_n(D^2/S^1) \cong \pi_n(S^2)$ are not that simple ! –  redfiloux Jun 2 '13 at 3:57
    
@redifloux: Thanks for pointing that out. This was something that was fuzzy in my thinking. I'd tacitly assumed that homotopy groups behave the same as homology groups for this case, but that's very far from true, as you point out! –  Grumpy Parsnip Jun 2 '13 at 7:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.