The property of suspension

Question

THe suspension $\Sigma X$ of a topological sapce $X$ is defined as the qutient space $$ \Sigma X=\dfrac{X\times [0,1]}{\sim}$$ Where $(x,t)\sim(y,s)$ if and only if $s=t=0$ or $s=t=1$ or $(x,t)=(y,s)$. Sow that $\Sigma X$ is simply connected if $X$ is path connected.

I hope to use Van kampen to prove this.In order to do that ,we can find two open covers whose fundamental groups are trivial and so does thier intersection.But I don't know how to find such open covers.

The statement you're trying to prove isn't true. Are you sure you've written it correctly? — Eric Wofsey, Nov 06 '16 at 02:18
Are you sure it's not that $X$ needs to be path connected? Otherwise, $X=S^0$ is a counterexample. — Dan Rust, Nov 06 '16 at 02:34

score 2 · Accepted Answer · answered Nov 06 '16 at 14:58

Let $U = (X \times [0,0.6))/{\sim}$ and let $V = (X \times (0.4,1])/{\sim}$. Note that $U$ and $V$ deformation retract onto a space which is homeomorphic to $CX$, the cone over $X$, which is always contractible. Now use Van Kampen's theorem to finish the proof (this is where you use path-connected).

The property of suspension

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