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Let $X$ be a connected and well pointed topological space (the latter means that $\{*\} \hookrightarrow X$ is a cofibration).

Is then $X$ already path connected? If yes why?

Background: Under the same conditions for $X$ as above I want to show that the reduced suspesion $\Sigma X$ is simple connected.

Using decompsion by cones $\Sigma X = C_+X \cup C_-X$ I intend to apply Seifert-van-Kampen. But here I need the path connectness of $X \cong C_+X \cap C_-X$.

user267839
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Unfortunately, the answer to your first question is "no." (You may have to assume path-connectedness for $X$ in order to get that $\Sigma X$ is simply-connected. I've never seen an argument for this that avoided making that assumption.)

Take as $X$ the topologist's sine curve with a basepoint $x_0$. If this is well-pointed, you already have your counterexample. If not, apply the "whisker trick" to $X$, where we attach an interval $[0,1]$ at $x_0$ and move the basepoint down to the dangling end of the interval. This resulting space is well-pointed, still connected, homotopy equivalent to $X$, but still not path-connected. This whiskered sine curve would then be a counterexample.

Randall
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  • But is the condition that $X$ path connected *neccessary* or only sufficent? Is there nevertheless a way to show that $\Sigma X$ is simply connected using the above conditions for $X$ despite it's not path connected? – user267839 Oct 21 '18 at 02:53
  • Well, that's a different question and I do not know the answer. – Randall Oct 21 '18 at 02:54