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
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
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 $\pi_1(S^2)$. |
|||
|
|
