What is an example of a Serre fibration that is not a Hurewicz fibration?
Asked
Active
Viewed 940 times
1 Answers
6
Let X be a mapping cylinder of the obvious map between the subspaces of the real line $\{n | n \in N\} \rightarrow {0}\cup\{1/n |n \in N^+\}$. Then the obvious map $X \rightarrow ({0}\cup\{1/n |n \in N^+\})\times I$ meets the requirement.
Denote $({0}\cup\{1/n |n \in N^+\})\times I$ by $Y$, and ${0}\cup\{1/n |n \in N^+\}$ by $Z$. So we have $f : X \rightarrow Y$.
Take the obvious map $k : Z \rightarrow X$. And an identity map $1_Y : Z\times I \rightarrow Y$ considered as a homotopy. Then the homotopy lifing fails.
On the other hand, it should not be hard to show that $f$ is a Serre fibration.
-
What should the codomain of that map be? Is it $\Bbb R$ or just the space ${1/n\mid n\in\Bbb N}$ ? – Stefan Hamcke Oct 21 '14 at 19:02
-
The codomain is ${1/n| n \in \mathbb{N}}\times I$. – Dimitri Chikhladze Oct 21 '14 at 20:05
-
Sorry, I meant the codomain of the first map you mention in your answer. – Stefan Hamcke Oct 21 '14 at 20:07
-
The codomain is {1/n|n∈N} and the comain is {n|n∈N}. Basically, you take the set of natural numbers with the discrete topology and the set of natural numbers with the topology of {1/n|n∈N}. Then you pair up their point and connect each of those pairs by lines. And that's X. – Dimitri Chikhladze Oct 21 '14 at 20:35
-
But the topology on ${1/n\mid n\in N}$ is the discrete, isn't it? – Stefan Hamcke Oct 21 '14 at 20:38
-
oh, I mean that set plus 0, which is an accumulation point. – Dimitri Chikhladze Oct 21 '14 at 20:41
-
I see. So you have an injection $X\to({0}\cup{1/n\mid n\in N})\times I$ which is not an embedding since the topology on the mapping cylinder $X$ is finer than the product topology. This $X$ is similar to the Arens-space. Thanks for clarifying. I think you should edit your answer to add the point $0$ to your set. – Stefan Hamcke Oct 21 '14 at 20:46
-
So you say this is not a fibration but a Serre fibration ... I will think about this. I'm currently studying different model structures on the category of topological spaces, and these structures differ by the type of fibrations that are involved, so this interests me :-) – Stefan Hamcke Oct 21 '14 at 20:51
-
Are you sure this is a Serre fibration? A path $h:I\to{0}\times I,h(t)=(0,t)$ does not have a lift in $X$. – Stefan Hamcke Oct 21 '14 at 21:02
-
Why? The lifting will also be $I \rightarrow {0}\times I, h(t)=(0,t)$. – Dimitri Chikhladze Oct 21 '14 at 21:13
-
Maybe I don't understand your mapping. Does your $N$ contain $0$ ? – Stefan Hamcke Oct 21 '14 at 21:15
-
It does, I use $N$ for natural numbers and $N^+$ for positive naturals. – Dimitri Chikhladze Oct 21 '14 at 21:18
-
And your mapping sends $0$ to $0$ and $n\ge1$ to $1/n$ ? – Stefan Hamcke Oct 21 '14 at 21:19
-
Yes, that is right – Dimitri Chikhladze Oct 21 '14 at 21:21
-
Oh, then I see my mistake. – Stefan Hamcke Oct 21 '14 at 21:22
-
2Ah, I understand why this is not a fibration. But it is a Serre fibration. In fact it has the HLP with respect to all connected spaces :-) – Stefan Hamcke Oct 21 '14 at 21:31