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If X is path connected how may i show that the reduced Suspension $\Sigma $ X is then simply connected. I cannot seem to picture this construction

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Do you know the seifert can kampen theorem? – mland Sep 13 '12 at 7:01
Whilst I thought it was true immediately by the Freudenthal suspension theorem, there are a serious of posts by Tom Goodwillie,, that suggests it might not (always be true). Although I can't see any flaw in the argument provided below... – Juan S Sep 13 '12 at 11:38
@JuanS: I haven't read the posts in detail, but in the first one the counterexample he produces involved the Hawaiian earring which does not satisfy the hypotheses of Freudenthal's suspension theorem. – Bruno Stonek Mar 10 '14 at 12:31

An overkilling answer could be the following: The Freudenthal suspension theorem tells us that, if $X$ is $n$-connected, then the natural morphism

$$ \pi_k(X) \longrightarrow \pi_{k+1}(\Sigma X) $$

is an isomorphism for $k\leq 2n$. Particularly, for $n=0$, we have an isomorphism $\pi_1(\Sigma X) = 1$.

But, if you want to "picture" the situation, take a look at this suspension drawing, and use the Seifert-van Kampen theorem, as mland points you. Particularly, look at Wikipedia's computation of $\pi_1(S^2)$.

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Well, this doesn't say why $\pi_0(\Sigma X)=0$ :) (which can be seen just from the definition of $\Sigma X$). – Bruno Stonek Jan 14 '15 at 16:22

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