Does simply connected implies locally pathwise-connected? 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 :)
 A: 

*

*Using the lifting criterion, why if $Y$ is simply connected, the lifting exists?


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.



*Does simply connected implies locally pathwise connected? And if it does, do you have a proof?


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.

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?

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 $V\subseteq U$.
