I am currently studying a little bit of covering spaces theory and I have a question.
Does simply connected implies locally pathwise-connected? I ask this because the lifting criterion states that a continuous map $f:Y\to X$, where $Y$ is connected and locally pathwise connected, exists iff $f_*(\pi_1(Y,y_0))\subset p_*(\pi_1(E,e_0))$. Now, I am trying to figure out why if $Y$ is simply connected this criterion holds. I was told that I have to prove that simply connected implies locally pathwise connected. I thought that this was trivial, since a simply connected space is pathwise connected I thought that this was locally true.
So what I am asking is:
- Using the lifting criterion, why if $Y$ is simply connected, the lifting exists?
- Does simply connected implies locally pathwise connected? And if it does, do you have a proof?
Also, since English is not my native language, I read both the terms locally path connected and locally pathwise connected, do these terms have the same definition or do they refers to different concepts?
Thank you for the help 🙂
Best Answer
This should be understood as "if $Y$ is additionally simply connected (to being locally path connected) then the lifting always exists". And that's because $\pi_1(Y)$ is trivial and thus $f_*(\pi_1(Y))\subseteq p_*(\pi_1(E))$ is always trivially satisfied.
Otherwise it is not necessarily true.
No, it doesn't. Take $A=\{\frac{1}{n}\ |\ n\in\mathbb{N}\}\cup\{0\}$ and then define $X=(A\times [0,1])\cup([0,1]\times \{0\})$. Also known as the comb space. Note that $X$ is not only simply connected, but even contractible. However it is not locally path connected.
These are the same things. And it means that for every point $x\in X$ and every open neighbourhood $U$ of $x$ there is a path connected open neighbourhood $V$ of $x$ such that $V\subseteq U$.