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
Simply connected reduced suspension on path connected X
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algebraic-topology
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0@JuanS: I haven'$t$ read $t$he 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. – 2014-03-10
1 Answers
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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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0Well, this doesn't say why $\pi_0(\Sigma X)=0$ :) (which can be seen just from the definition of $\Sigma X$). – 2015-01-14