When is the free loop space simply connected? I am not sure if there is an obvious answer to this, but this has been bothering me.
Let $X$ be a topological space. When is the free loop space, $LX$, simply connected?
Correct me if I'm wrong, but I believe that if $X$ is simply connected, then $LX$ is also simply connected since if we take a loop of loops, we can always contract it to a point. Is this an if and only if statement? Is there an example of a space that is not simply connected but its loop space is?
 A: First let's look at the simpler case of the based loop space $\Omega X$ (pick a basepoint). WLOG $X$ is connected since $\Omega X$ only sees the connected component containing the basepoint. This case is simpler because we know that
$$\pi_i(\Omega X) \cong \pi_{i+1}(X)$$
so we conclude that


*

*$\Omega X$ is connected iff $X$ is simply connected, and

*$\Omega X$ is simply connected iff $X$ is $2$-connected ($\pi_1, \pi_2$ both vanish).


In particular, $\Omega X$ is always less connected than $X$ is; taking based loops pulls all of the homotopy groups down one index. 
Now, the free and based loop spaces can be related by a fibration sequence
$$\Omega X \to LX \to X$$
which gives us a long exact sequence in homotopy the relevant part of which is
$$\cdots \to \pi_1(\Omega X) \to \pi_1(LX) \to \pi_1(X) \to \pi_0(\Omega X) \to \pi_0(LX) \to \pi_0(X) \to \cdots$$
This shows in particular that a sufficient condition for $\pi_0(LX)$ and $\pi_1(LX)$ to vanish is that $\pi_0(X), \pi_1(X), \pi_2(X)$ all vanish; that is, if $X$ is $2$-connected, then $LX$, like $\Omega X$, is simply connected. However, I don't know if the converse is true.
The simplest case to understand is when $X = G$ is a topological group; in that case, the fibration sequence trivializes to a product decomposition
$$LG \cong G \times \Omega G$$
(as spaces, not as topological groups!) which shows in particular that
$$\pi_0(LG) \cong \pi_0(G) \times \pi_1(G), \pi_1(LG) \cong \pi_1(G) \times \pi_2(G).$$
So in this case $LG$ is simply connected iff $G$ is $2$-connected, and in particular it is not enough to require that $G$ is simply connected. (However, if $G$ is a simply connected compact Lie group, then it turns out that this automatically implies that $G$ is $2$-connected.) 
A: No, it is not true. An example is the sphere $S^2$: it is simply connected, but its loop space is not. Take a point $x \in S^2$ and the loop of loops that has the following properties:


*

*the two endpoints are two loops which have constant value $x$;

*at every time it is a loop based on $x$;

*as the time passes, it "spans" the whole sphere.


More formally, if you see the sphere $S^2$ as $[0,1] \times [0,1]$ with the boundary contracted to a point, take the map $(s, t) \mapsto [(s,t)]$. This is not trivial in the loop space of $S^2$.
More in general, iterating the loop space construction on a pointed topological space $(X, x)$ one obtains the so-called higher homotopy groups $\pi_i(X, x)$. For every $i$ there holds $\pi_i(X, x) = \pi_{i-1}(\pi_1(X, x), \gamma_x))$, where $\gamma_x$ is the loop that takes the constant value $x$,
