locally simply-connected vs. semilocally simply-connected Can someone help me understand the difference between locally simply-connected and semilocally simply-connected?
I actually need more help understanding (with an example), what it means to be locally simply-connected. 
I know that a space X is semi-locally simply connected if every point in X has a neighborhood U for which the homomorphism from the fundamental group of U to the fundamental group of X, induced by the inclusion map of U into X, is trivial. And I understand why the Hawaiian earring is NOT semilocally simply-connected. 
Another example that I was trying to understand was that "$CX = (X \times I) / (X \times \{0\})$ (where $X$ is the Hawaiian earring) is semi-locally connected since its contractible, but its not locally simply-connected" - I didn't get why its not locally simply-connected as I don't know how to check for locally simply-connectedness.
edit: My understanding now is that a part of the intermediate loop in the homotopy is permitted to be outside of U, for a semilocally simply-connected space, but cannot be allowed for a locally simply-connected space. But my question about CX remains.  
 A: First I would like to point out you say the Hawaiian Earring $X$ is semilocally simply connected when it is not. Any neighborhood of the limit point will contain a full circle (infinitely many but we only need one) and the inclusion will send the generating loop for that circle to the same generator of the fundamental group of $X$ thus not the trivial map.
Now to start, let's state the definition of locally simply connected

A space $X$ is locally simply connected if for every $x \in X$ and every neighborhood $V$ of $x$ there is a neighborhood $U \subset V$ of $x$ that is simply connected in the subspace topology.

So take the cone $CX$ of the Hawaiian Earring and examine neighborhoods of the point $\{x\} \times \{1\}$ where $x$ is the limit of the shrinking circles in $X$. Any small neighborhood $U$ of this point deformation retracts onto a neighborhood of $x$ in $X$ (it looks like $U \times I$ maybe with some of the sides shortened but that doesn't change anything homtopically) and thus $U$ is not simply connected. So we can conclude $CX$ is not locally simply connected.
What makes the difference for semilocally simply connected is you can any neighborhood in $CX$, doesn't matter what the fundamental group is and we have the inclusion $$i:\pi_1(U) \hookrightarrow \pi_1(CX) = 0$$
must be the trivial map.
