How to compute $\pi_0$ of $Maps(S^1, \Omega^2({S}^2, p))$ Denote by $\Omega^2({S}^2)$ the space of DOTTED maps from the $2-$sphere $S^2$ onto itself. And consider its FREE loop space $X=\mathcal{L}(\Omega^2({S}^2))=Maps(S^1, \Omega^2({S}^2))$. I think that $\pi_0(X)$ is $\pi_3(S^2)\oplus\pi_2(S^2)$ where the integer in $\pi_2(S^2)$ measures the degree of the map $\{\cdot\}\times S^2\to S^2$, $\{\cdot\}\in S^1$. But i need to know the geometric meaning of the term in $\pi_3(S^2)$ there; i.e. if a given element has zero the element in $\pi_2(S^2)$, how do we recognize visually or geometrically the other integer? Thanks in advance 
 A: I will assume that "dotted" means the same as "basepoint-preserving". 
There are homeomorphisms $$\mathcal L\Omega^2 S^2\cong \mbox{map}_*(S^1_+\wedge S^2, S^2)\cong \mbox{map}_*(S^3/S^1, S^2).$$ Note that there is a homotopy equivalence $S^1_+\wedge S^2\simeq S^3\vee S^2$. Therefore, there is a homotopy equivalence $\mathcal L\Omega^2 S^2\simeq \Omega^3S^2\times \Omega^2S^2$. From here it follows that $$\pi_0(\mathcal L\Omega^2 S^2)\cong \pi_3(S^2)\oplus \pi_2(S^2),$$ exactly as you said. 
Elements of $\pi_2(S^2)\cong {\mathbb Z}$ correspond to the degree of the map $S^2\to S^2$, still exactly as you said. On the other hand, elements of $\pi_3(S^2)\cong \mathbb Z$ correspond to the Hopf invariant of a map $S^3\to S^2$. 
Suppose $f\in\mathcal L\Omega^2 S^2$. Interpret $f$ as a map $f\colon S^3/S^1\to S^2$. Let $g\colon S^3\to S^2$ be the composition of $f$ with the obvious quotient map. The Hopf invariant of $g$ is the answer to your question. 
The Hopf invariant has several equivalent definitions. The most common one uses the cup product on the mapping cone of a map. The simplest geometric one is probably the following: choose points $x, y$ in $S^2$ such that the preimage of each point in $S^3$ is a disjoint union of circles. This can be achieved by approximating $g$ with a differentiable function and choosing two regular values. Then $$H(g)=\mbox{Lk}(g^{-1}(x), g^{-1}(y)).$$ 
(Lk is the linking number)
