Is there any closed embedding which is not cofibration? I firstly think that if $X$ is Topologist's sine curve and $A$ is $(0,0)$, then embedding $i:A\rightarrow X$ might satisfy this condition. However, I couldn't prove there is no retraction $r:X\times I\rightarrow (X\times 0)\cup(A\times I) $.
Asked
Active
Viewed 265 times
1 Answers
4
Let $X=[0,1]$ and let $A$ be the Cantor set inside $X$. Then $i:A\to X$ is a closed embedding but there is no retraction, even continuous surjection $X\times I\to X\times\{0\}\cup A\times I$ because $X\times\{0\}\cup A\times I$ is not a Peano space (it is not locally connected) while $X\times I=[0,1]^2$ clearly is. I'm referring to the Hahn–Mazurkiewicz theorem.
This can be generalized to any pair $(A, X)$ with $X$ compact and locally connected, while $A$ closed in $X$ but not locally connected.
freakish
- 47,446