# Suspension of Connected Space

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.

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.