Simply connected reduced suspension on path connected X 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
 A: 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 $\pi_1(S^2)$.
A: Here's a brief outline:
Note that $\Sigma X = X \times I / ( X \times \partial I \cup \{x_0\} \times I )$, where $x_0$ is assumed to be the base point of $X$ and $\partial I = \{0,1\}$.
Taking any loop $f: [0,1] \to \Sigma X$, one can define the following homotopy:
$$
H(x, t) = [(f(x), 1-t)]
$$
where for any $\alpha \in X \times I$, $[\alpha]$ denotes its equivalence class in $\Sigma X$.
Use the universal property of quotient spaces to show that this map is continuous, and check that it is also a homotopy. This will show that every loop is homotopic to the constant loop, and thus, $\pi_1(X) = 0$.
As for connectedness of $\Sigma X$, just note that $X$ is connected, so $X \times I$ is connected, and quotients of connected spaces are connected, therefore $\Sigma X$ is connected.
