A continuous map $f: S^1 \to S^2$ is homotopic to a point $f: S^1 \to S^2$ is a continuous map. Prove that $f$ is homotopic to a constant function.  Can someone give me a dircetion for proving this result? I think I should use the fact that $S^2$ is simply connected. I was able to prove the result given that $Im(f) \neq S^2$, but this is not generally true.
 A: You can use the Lebesgue number lemma (though this may be overkill in the specific case of a sphere).
Specifically, take the open cover $\mathcal{O}=\{H^+,H^-\}$ of $S^2$ consisting of two thickened hemispheres. The Lebesgue lemma gives some number $\varepsilon$ such that any $\varepsilon$-ball in $S^2$ is contained in one or the other element of $\mathcal{O}$. Because $f$ is continuous on the compact set $S^1$, $f$ is uniformly continuous; choose $\delta$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$.
Now, divide $S^1$ into intervals $I_1,I_2,\dots,I_n$ of radius less than $\delta$, with midpoints $p_1,p_2,\dots,p_n$. Then $f(I_k)$ is always contained entirely within the $\varepsilon$-ball centered at $f(p_k)$, and hence within some element of $\mathcal{O}$. But any path in the thickened hemispheres $H^+$ and $H^-$ is homotopic (without moving the endpoints) to a great circle path, and so $f|_{I_k}$ is homotopic to a great circle path (without moving the endpoints).
It follows that $f$ is homotopic to some loop $g$ which is the union of finitely many great circle paths. But $g$ is nullhomotopic, as you've noted, and so $f$ is as well.
A: Use Van Kampens theorem for $S^2$ with open cover containing the upper hemisphere and lower hemisphere. Then stretch these two open sets a bit so that they both cover the equator. Use Van Kampens theorem from here, you know these two open sets are simply connected so by Van Kampens theorem we know that $S^2$ is simply connected aswell. This argument works for every sphere $S^n$ except $S^1$ since in that case the equator isn't path connected.
A: 
This lemma can be used. See the proof if you are interested.
( $h_*[f]=[hof]$)

Now since $S²$ is simply connected $h_*$ is trivial and hence h is null homotopic
