1
$\begingroup$

Let $(X,A)$ be a "good" CW pair. Let $*\in A \subset X$ be the base point of $ X$ and $CA$ the cone on $A$ i want to show that $(X\cup CA)/CA$ is homeomorphic to $X/A$. I can see it geometrically but i want to prove it. Consider the composite $f:X\cup CA \to X \to X/A $ where the first map sends $x$ to $x$ and $(a,t)\in CA$ to $a$ and the second map is the quotient map. $f$ is a surjective map sending $CA$ to the base point of $X/A$ which is the class of $*$ identified with $A$ so $f$ factorizes through $(X\cup CA)/CA$ and the induced map

${\tilde f}:(X\cup CA)/CA\to X/A$ is a homeomorphism. Is this correct?

  • 0
    i $c$hanged it.. thanks2011-11-28

1 Answers 1

1

Following your argument you only get a continuous bijection and you will have to prove that your map's inverse is continuous.

You can see this by doing the same thing of what you did from the other direction with the map $g: X \rightarrow X \cup CA \rightarrow (X \cup CA) /CA $ which is inclusion followed by the quotient map, hence continuous. Obviously it is surjective. As in your argument, $g(A)$ is the class of $*$, so you get an induced map $\tilde{g}: X/A \rightarrow (X \cup CA) /CA$ and that is your continuous inverse.

It seems to me that $(X,A)$ does not have to be a CW pair for this to work.

  • 0
    i had the example of collapsing a contractible subspace in mind. As you can see collapsing $S^1-*$ in $S^1$ is not of the homotopy type of $S^1$ so that's why i used the Cw pair here but i think you are right it is irrelevant here.2011-11-28