4

Let $X$ be the union of all circles centered at $(0,\frac{1}{n})$ with radius $\frac{1}{n}$ for $n\in N$. Let $A$ be $(0,0)$. Show that $A\to X$ is not a cofibration.

This appears as a non-example for inclusion != cofibration. However I am not able to show it.

Ma Ming
  • 7,482

1 Answers1

4

Hint: The inclusion $i\colon A\hookrightarrow X$ is a cofibration if and only if $M(i)$ is a retract of $X\times I$, where $M(i)$ is the mapping cylinder $(X\times\{0\})\cup(A\times I)$.

Consider the implications of such a continuous retract existing for the inclusion of the base point into the Hawaiian earrings, and in particular that the existence of such a map gives a contradiction. (Don't be afraid to get down and dirty with open sets - these spaces all embed nicely into $\mathbb{R}^3$ so your intuition should be used as much as possible.)

Dan Rust
  • 30,108
  • Suppose, to the contrary, that $f\colon X\times I\to X\times {0}\cup A\times I$ is a retract. Since $f$ is continuous (uniform continuous), for any $\varepsilon>0$ there exist $\delta>0$ such that, if $|x-y|<\delta$ then $|f(x)-f(y)|<\varepsilon$. Pick $x=(0,0,1)$, images of $X_n\times{t}$ are a coninuous family of circles in $X\times {0}\cup A\times I$. Since $X_n\times{0}$ is fixed, the image of $X_n\times{0}$ is at least $A\times I\cup X_n\times {0}$. As $n$ go to infinity, we obtain a contradiction to the previous continuous condition. – Ma Ming Nov 26 '13 at 11:32