# Exact sequence for homology of space, subspace and quotient space

Hatcher states the following theorem on page 114 of his Algebraic Topology:

If $X$ is a space and $A$ is a nonempty closed subspace that is a deformation retract of some neighborhood in $X$, then there is an exact sequence $$...\longrightarrow\widetilde{H}_n(A)\overset{i_*}\longrightarrow \widetilde{H}_n(X)\overset{j_*}\longrightarrow\widetilde{H}_n(X/A)\overset{\partial}\longrightarrow \widetilde{H}_{n-1}(A)\overset{i_*}\longrightarrow...$$

where $i: A\hookrightarrow X$ is the inclusion and $j:X \rightarrow X/A$ is the quotient map.

Perhaps I am having a brain malfunction at the moment, but what are some interesting nonempty spaces which do not satisfy these criterion? By interesting, I mean something that appears "in nature."

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If by interesting space you mean CW-complex, and by subspace you mean subcomplex (and this would not be absurd!) then no, there no examples. –  Mariano Suárez-Alvarez Nov 18 '12 at 20:54

## 3 Answers

If you're looking for an arbitrary topological space with this property, take $X = \{a,b,c\}$ and let the topology $\mathcal{T} = \{\emptyset, \{a,b,c\},\{c\},\{a,c\},\{b,c\}\}$. Consider the closed subset $C = \{a,b\}$. The only open set that might retract onto $C$ is the whole space $X = \{a,b,c\}$. Let us suppose that $r(a) = a, r(b) = b, r(c) = b$. Then the inverse of the open set $\{a,c\}$ is just $\{a\}$, which is not an open set, so $r$ is not continuous.

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One example which occurs a lot "in nature" is when $A$ is a large open subset of $X$. For example, $X$ is a manifold and $A = X - \text{point}$ or when $X$ is the total space of a vector bundle and $A$ is the complement of the base space zero section. However, in these cases, it is often possible to "shrink" A so that the condition remains true.

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The complement of the zero section, rather ;-) –  Mariano Suárez-Alvarez Nov 18 '12 at 20:55
Indeed, thank you, correcting ... –  apurv Nov 19 '12 at 4:54

Remember that to be a deformation retract of a neighbourhood, you have to be locally path connected. If you think about this, you'll see that the comb space gives you an example. To be precise, if $X$ is the space $[0,1]\times\{0\}\cup(\{\frac{1}{n};n\in\mathbb{N}\}\cup\{0\})\times [0,1]$ and $A$ is $\{0\}\times[a,1]$ for some $a>0$, then $A$ is closed, but no neighbourhood deformation retracts onto it. You can find plenty of examples based on this.

If you're wondering why the $a>0$ is there, it's because I'm not sure of Hatcher's terminology. What he calls deformation retracts I call strong deformation retracts. I think he calls my deformation retracts weak deformation retracts in an exercise somewhere. Note that $\{0\}\times[0,1]$ is a weak deformation retract of $X$, and I think the proof of the theorem you quote goes through with this weaker assumption.

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