Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello !

Let $X$ be a locale or a topological spaces. $I$ denote the unit interval of the real numbers, and $X^I$ the space of function from $I$ to $X$ (The locale exponential if $X$ is a locale or the set of function endowed with the open-compact topology if $X$ is a topological space.)

In both case, we consider the map $X^I \rightarrow X \times X$, which is simply the evaluation at the two endpoints.

In "Connected locally connected toposes are path connected" (available here : http://www.ams.org/journals/tran/1986-295-02/S0002-9947-1986-0833712-3/) Moerdijk and Wraith showed that if $X$ is a locale (or actually even a topos) which is connected and locally connected then the map $X^I \rightarrow X \times X$ is an open surjection.

At the very end of the paper, they mentioned without proof that in the case of topological space, the map $X^I \rightarrow X \times X$ is open if and only if $X$ is "semi-locally path connected".

But if I assume that $X^I \rightarrow X \times X$ is an open map, and $U$ is any open set of $X$, then $U^I$ is an open of $X^I$. hence by restriction, the map $U^I \rightarrow U \times U$ is also an open map. In particular its image which is the relation "$x$ is path connected with $y$ in $U$ " is open in $U \times U$.

So if $x$ is any point of $U$ the path connected component of $x$ in $U$ is the set of $y$ path connected to $x$ in $U$, and hence is open by the previous observation. Finally, any open of $X$ has open path connected component, ie $X$ is locally path connected.

I don't exactly know what is meant by "semi-locally path connected" but I though it was weaker than "locally path connected" so there is something weird here !

My questions are :

  • Is there a mistake in what I just said, a mistake in the paper, or am I misunderstanding the notion "semi-locally path connected " ?

  • For a general locale, doesn't the openness of the map $X^I \rightarrow X \times X$ actually equivalent to the fact that $X$ is locally connected ?

Thank you !

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Here are what I believe are the standard definitions:

A topological space $X$ is locally path-connected if for every $x\in X$ and every open neighbourhood $U$ of $x$, there exists a smaller open neighbourhood $V\subseteq U$ of $x$ such that $V$ is path-connected. Equivalently, path components of open subsets are open.

A topological space $X$ is semi-locally path-connected if for every $x\in X$ and every open neighbourhood $U$ of $x$, there exists a smaller open neighbourhood $V\subseteq U$ of $x$ such that every $y\in V$ is connected to $x$ by a path whose image lies in $U$ (but not necessarily in $V$).

So semi-locally path-connected is formally weaker. But in fact, the definitions turn out to be equivalent for all spaces (this is an exercise on this sheet).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.