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$(X,A)$ is a good pair if $\exists V\subset X$ s.t. $V$ is a neighbourhood of $A$ that deformation retracts to $A$.

Prove that if $(X,A)$ is a good pair, then $(X/A,A/A)$ is also a good pair. How about the converse?

It seems to me that $A/A$ ends up being just a point in the quotient space of $X/A$ and so we should be able to find a neighbourhood of $A/A$ ($V/A$ perhaps?) that deformation retracts to $A/A$. What do I need to do to finish up the argument (if I'm in the right direction that is...)

I'm guessing that the converse is not true. I'm thinking something along the lines of $X$ being the Hawaiian earring but when quotient by a particular $A$, it becomes "nice".

Stefan Hamcke
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Haikal Yeo
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1 Answers1

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$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} $$

The homotopy $h$ from Id$_V$ to $r$ gives rise to the commutative diagram \begin{array}{ccc} V\times I & \ra{h} & V \\ \da{q\times Id} && \da{q}\\ V/A\times I & \ra{\tilde h} & V/A \end{array} since $q\times$Id is a quotient map as $I$ is locally compact. Then it is easy to verify that $\tilde h$ is a homotopy from Id to $\tilde r$ which is constant on $A/A$.

The converse need not be true: Take $X=I\cup\{0'\}$, a unit interval with two origins, where the neighborhood base of $0'$ is formed by intervals $[0',\epsilon):=\{0'\}\cup(0,\epsilon)$. Let $A=\{0,0'\}$. Then $X/A$ is homeomorphic to $I$ with $A/A$ the origin, which is clearly a good pair. But $(X,A)$ is not a good pair since $A$ is not even a retract of some neighborhood.
Note that this counterexample makes explicit use of the fact that both origins cannot be separated by disjoint neighborhoods, and I don't know of a counterexample for Hausdorff spaces.

Stefan Hamcke
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