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

How to show that in this case the reduced suspesion $\Sigma X$ is simple connected.

Remark: This question result from a former thread of mine: Suspension of Connected Space Simply Connected

As @Randall explained we can't expect that $X$ is path connected in generally. But my question is if the statement that the reduced suspesion $\Sigma X$ is simple connected is nevertheless true.

user267839
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1 Answers1

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It is not true. Take any connected but not pathwise connected well-pointed $(X,*)$. Since $(X,*)$ is well-pointed, the reduced suspension has the same homotopy type as the unreduced suspension $SX$. We shall show that $SX$ is not simply connected which implies that $\Sigma X$ is not simply connected.

Working with reduced singular homology, we know that $H_1(SX) = \tilde{H}_1(SX) \approx \tilde{H}_0(X)$. See Reduced Homology on unreduced suspension . Since $X$ is not pathwise connected, $H_0(X)$ is a free abelian group with more than one generator, hence $\tilde{H}_0(X) \ne 0$ and thus $H_1(SX) \ne 0$.

The first homology group is the the abelianization of the fundamental group (see The First Homology Group is the Abelianization of the Fundamental Group.}). Hence the fundamental group of $SX$ cannot be trivial.

Paul Frost
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