# Why is this map $S^2\to S^1$ nullhomotopic?

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.

• 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. – D. Thomine Mar 29 at 10:56
• 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. – D. Thomine Mar 29 at 10:58
• @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 $M$. – Javi Mar 29 at 11:45
• 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" $H^1(-;\mathbb{Z})$. – Connor Malin Mar 29 at 16:26
• Thanks for your comment @ConnorMalin That's very intersting! – Javi Mar 29 at 17:30

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.

• By the closed paht $q$ you mean the map $\phi$ on my question? – Javi Mar 29 at 11:36
• 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 $q : [0,1] \to S^1$. – Paul Frost Mar 29 at 11:40
• 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? – Javi Mar 29 at 11:42
• It is continuous because multplication $\mu : [0,1] \times [0,1] \to [0,1], \mu(s,t) =st$, is continuous. – Paul Frost Mar 29 at 11:44
• 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. – Paul Frost Mar 29 at 11:50

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.

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.