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Consider a (EDIT: Hausdorff) topological space $X$ in which for every two points $x,y\in X$ there exists a "bridge" point $z\in X\setminus\{x,y\}$ such that $x$ and $y$ belong to different connected components of $X\setminus z$.

Is the fundamental group of $X$ trivial?

In my specific case $X$ is path-connected (and thus, I guess, arc-connected), does this help?

EDIT: In my specific case, the space is Hausdorff. At that moment I thought it was implied by the property, but now I see it should be specified.

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  • $\begingroup$ For $S^2$ the fundamental group is trivial., For $S^1$ it's not. These are two basic examples. $\endgroup$ – Orest Bucicovschi Sep 12 '17 at 2:09
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    $\begingroup$ @orangeskid I don't see how either of those satisfy the property "for every two points $x,y\in X$ there exists a point $z\in X\setminus\{x,y\}$ such that $x$ and $y$ belong to different connected components of $X\setminus \{z\}$." $\endgroup$ – anon Sep 12 '17 at 2:11
  • $\begingroup$ OP: do you have any examples of such spaces handy (besides things homeo to intervals)? $\endgroup$ – Randall Sep 12 '17 at 2:12
  • $\begingroup$ Oh, I see, you want spaces with that property... apart from intervals... not sure then $\endgroup$ – Orest Bucicovschi Sep 12 '17 at 2:13
  • $\begingroup$ @Randall Any tree (1-complex). Actually my space is pretty much like a tree: the point $z$ has a 1-dim neighborhood, i.e., is on an "edge" of the "graph". But there can be infinitely many such "edges", and the topology can be more complicated than just a tree. $\endgroup$ – Alexander Gelbukh Sep 12 '17 at 2:15
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For Hausdoff spaces, the answer is yes, $X$ is simply connected. The answer here uses some continuum theory. The key is to notice that $X$ is uniquely arcwise connected.

First notice that you cannot embed $S^1$ into $X$ because of your required property. This means that $X$ contains no simple closed curves (homeomorphic images of $S^1$). It is well-known that a Hausdorff space $X$ is uniquely arcwise connected if and only if it does not contain a simple closed curve (this might be an exercise in a continuum theory book, e.g. Nadler's book). So $X$ must be uniquely arcwise connected.

Now let $\alpha:(S^1,p)\to (X,x)$ be a loop based at $p$. The image of $S^1$ in a Hausdorff space is a Peano continuum (a path-connected, locally path connected, compact metric space). Moreover $\alpha(S^1)$ must be uniquely arcwise connected since it is a path-connected subspace of a uniquely arcwise connected space. But any uniquely arcwise connected Peano continuum is a dendrite and all dendrites are contractible. Since $\alpha$ factors through a contractible space, it is null-homotopic.

For non-Hausdorff spaces, path connected does not even imply arcwise connected so the result could potentially be false.

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