# Space with semi-locally simply connected open subsets

A topological space $$X$$ is semi-locally simply connected if, for any $$x\in X$$, there exists an open neighbourhood $$U$$ of $$x$$ such that any loop in $$U$$ is homotopically equivalent to a constant one in $$X$$ or, equivalently, if the functor $$\Pi_1(U)\rightarrow\Pi_1(X)$$ induced by the inclusion $$U\subseteq X$$ factorizes through a groupoid in which for each pair of objects there is at most one morphism.

My question is: is it true that if a locally path connected space $$X$$ is such that, for any open subset $$U\subseteq X$$, $$U$$ is semi-locally simply connected, then $$X$$ must be locally simply connected?

• Sorry, I've misread. – Saucy O'Path Oct 21 '18 at 16:29
• Your equivalent reformulation of "any loop in $U$ is homotopic to a constant one in $X$" is only valid if $U$ is pathconnected : you should say that the image of this functor has at most one arrow between any two points – Max Oct 23 '18 at 9:32
• You're right, in my mind I was assuming without a reason that $X$ is locally path connected. Thank you, I will make an edit. – mfox Oct 23 '18 at 9:42
• @user126154 The two-point discrete topology is not simply connected (since it's not path connected), yet still every open subset is semilocally simply connected. – Kyle Miller Dec 20 '18 at 18:52
• @user126154 Every open set of a manifold is semilocally simply connected (since manifolds are locally simply connected), but many manifolds aren't simply connected. – Kyle Miller Dec 21 '18 at 20:17

Consider the subspace of $$\mathbb{R}^2$$ that is the union of line segments from $$(0,0)$$ to $$(x,1)$$, where $$x\in\{0\}\cup\{1/n:n\in\mathbb{Z},n>0\}$$. This is simply connected and "locally semilocally simply connected", but it is not locally simply connected because it is not locally path connected. (The definition of simply connected is that $$\pi_0$$ and $$\pi_1$$ are trivial.)
Perhaps the hypotheses imply that the space is locally $$\pi_1$$-trivial -- I could not think of an example that wasn't.
• $CX$ is semilocally simply connected as you say, but it's not true that each one of its open subset is semilocally simply connected ($X\times [0, 1)$ for example is not semilocally simply connected), so $CX$ is not a valid counterexample. – mfox Oct 23 '18 at 8:35
• @mfox I misunderstood your question because of the re-use of the letter $U$ (I should have read it more carefully!) I've replaced the answer, though I'd much rather have a counterexample that doesn't feel like it's getting by on a technicality. – Kyle Miller Oct 30 '18 at 14:38