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I know that $\pi_2(S^1)=0$ since $S^1$ has $\mathbb{R}$ as universal cover, which is contractible. However, I have a map $S^2\to S^1$ that I can't intuitively see why it is nullhomotopic (it has to be, since otherwise it would represent a nontrivial element of $\pi_2(S^1)$).

Take $S^2$ to be the standard sphere centered at the origin of $\mathbb{R}^3$. Project onto the $XY$-plane by $p_1: (x,y,z)\mapsto (x,y,0)$. Then, do the same with the disc obtained, $p_2:(x,y,0)\mapsto (x,0,0)$. Now we have an interval, that we can send homeomorphically (say by $h$) to $[0,1]$. Now, I can choose non-nullhomotopic maps $[0,1]\to S^1$, such as the quotient map $q$ (not nullhomotopic because it represents a generator of singular homology) or $\phi(t)=e^{2\pi i t}$ (not nullohomotpic because it represents a generator of the fundamental group).

All the maps are continuous. The resulting map $f=q\circ h\circ p_2\circ p_1$ (or $g=\phi\circ h\circ p_2\circ p_1$) should be nullhomotopic, but I can't figure out a homotopy or a geometric intuition of how is that possible, since what I see is that in the end we're just performing the classical loop around $S^1$, which is not nullhomotopic.

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    $\begingroup$ So your problem is simpler. Any map $[0,1] \to \mathbb{S}^2$ is nullhomotopic. If you quotient the endpoints of $[0,1]$, then you can get a non-nullhomotopic map, but then the composition doesn't work out. $\endgroup$ – D. Thomine Mar 29 at 10:56
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    $\begingroup$ So, essentially, what you've proved is that, if you identify (quotient) $(1, 0, 0)$ with $(-1,0,0)$ on the sphere, then there is a non-nullhomotopic map from the resulting space $M = \mathbb{S}^2_{/\sim}$ to $\mathbb{S}_1$. Which is true. But $M$ is not a sphere. $\endgroup$ – D. Thomine Mar 29 at 10:58
  • $\begingroup$ @D.Thomine I don't understand what you mean by "then the composition doesn't work". I defined a map $S^2\to S^1$. In fact, my first idea was defining it passing first to $M$, but it was more difficult to described. But, in that case, it is just a map $S^2\to S^1$ which factors through $M$. $\endgroup$ – Javi Mar 29 at 11:45
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    $\begingroup$ There are a great many answers here- it shows that there are an incredible number of ways to show maps into $S^1$ are nullhomotopic. There is a generalization of the covering space argument that shows that if $\pi_1 (X)$ has only the trivial homomorphism into $\mathbb{Z}$, then any map into $S^1$ is nullhomotopic. If you are familiar with cohomology. this also implies that this happens if and only if $H^1(X;\mathbb{Z})$ is trivial. You might wonder then if this can be proved with cohomology, and it can! This is because $S^1$ "represents" $H^1(-;\mathbb{Z})$. $\endgroup$ – Connor Malin Mar 29 at 16:26
  • $\begingroup$ Thanks for your comment @ConnorMalin That's very intersting! $\endgroup$ – Javi Mar 29 at 17:30
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You claim that $q : [0,1] \to S^1$ is not nullhomotopic. But it is. Define $h : [0,1] \times [0,1] \to S^1, h(s,t) = q(st)$. Then $h(s,0) = q(0)$ which is constant and $q(s,1) = q(s)$.

You misunderstanding comes from the fact that the closed path $q$ is a generator of $\pi_1(S^1)$. But for fundamental groups we consider homotopies of closed paths which keep the endpoints $\{ 0, 1 \}$ of $[0,1]$ fixed. This kind of homotopy is a very special one. For maps $S^2 \to S^1$ we are allowed to use arbitary homotopies. Even if we require that some basepoint of $S^2$ is kept fixed under homotopies ("pointed homotopies" ), we get the same result: All pointed maps are pointed homotopic to the constant pointed map.

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  • $\begingroup$ By the closed paht $q$ you mean the map $\phi$ on my question? $\endgroup$ – Javi Mar 29 at 11:36
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    $\begingroup$ Yes. In fact, your map $\phi$ is a quotient map $[0,1] \to S^1$. I would even say it is the standard quotient map. But we can take any quotient map $q : [0,1] \to S^1$. $\endgroup$ – Paul Frost Mar 29 at 11:40
  • $\begingroup$ So, in the same way we have the homotopy $h$, wouldn't we have $H:[0,1]\times [0,1]\to S^1$, $H(s,t)=\phi(st)$? $\phi(s,0)=\phi(0)$ which is constant and $\phi(s,1)=\phi(s)$. Does it fail to be continuous? $\endgroup$ – Javi Mar 29 at 11:42
  • $\begingroup$ It is continuous because multplication $\mu : [0,1] \times [0,1] \to [0,1], \mu(s,t) =st$, is continuous. $\endgroup$ – Paul Frost Mar 29 at 11:44
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    $\begingroup$ As I explained in my answer, for fundamental groups we need homotopies keeping $\{ 0,1\}$ fixed. The above homotopy only keeps $0$ fixed, thus it is not a homotopy of paths. But nevertheless $\phi$ is nullhomotopic. $\endgroup$ – Paul Frost Mar 29 at 11:50
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It may not be easy to describe a nullhomotopy on the end result, but you have intermediate steps (a disc and a line segment) which are trivially nullhomotopic. Insert a homotopy at one of those points in your function chain.

The image of the resulting homotopy may look discontinuous, as you're tearing a hole in the circle. But the parts that are torn apart do not correspond to points close together on $S^2$, so it is, in fact, still continuous.

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Imagine poking the sphere inwards from $x=0$ and $x=1$ so that under projection to the $x$-axis it doesn't reach all the way around the interval $[0,1]$. Performing this in $\mathbb R^3$ exhibits a homotopy between two embeddings of $S^2$. From this point it should be clear that following this homotopy with the map you describe yields a nullhomotopic map.

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