degree 1 map $f : M \to S^n$

Question

Let us consider $M$ a closed and connected n-manifold which is also orientable. An exercise in Hatcher claims that for any such $M$ there is a continuous map $f: M \to S^n$ such that it's degree is 1, i.e. the induced map $f_*: H_n\left(M\right) \to H_n\left(S^n\right)$ sends the fundamental class $\left[M\right]$ to $\left[S^n\right]$. My idea was the following:

Consider an arbitrary point $x \in M$. Then take a chart $\phi: U_x \xrightarrow{\approx} \mathbb{R}^n$ that maps $x$ to a nonzero value $\phi(x)$. Then normalize this value to $\frac{\phi(x)}{\Vert \phi(x) \Vert}$. We define our map $f: M \to S^n$ to be exactly this map. Does this construction work? I think from here on I'd have a solution to the exercise but I don't know if this map $f$ is valid.

This construction is highly dependent on the chosen chart, and thus does not (obviously) extend to a global mapping. — Mees de Vries, May 20 '16 at 11:31
Absolutely. But is it even possible my making a smart choice? — noctusraid, May 20 '16 at 11:45
No that doesn't work. Your map would be to $S^{n-1}$ anyway! — Cheerful Parsnip, May 20 '16 at 11:50

score 11 · Answer 1 · answered May 20 '16 at 11:52

11

My favorite method of constructing a degree $1$ map is to take an embedded open ball in $M$ and crush its complement to a point.

answered May 20 '16 at 11:52

Cheerful Parsnip

28,251

score 6 · Accepted Answer · answered May 20 '16 at 13:51

6

Theorem:For a map $f:M\to N$ between connected closed oriented $n-$manifolds, suppose that there is a ball $B\subset N$ such that $F^{-1}(B)$ is the disjoint union of balls $B_i$ each mapping homeomorphically onto $B$. Then $deg(f)=\sum \epsilon_i$ where $\epsilon_i =\pm 1$ depending on whether $f|B_i$ preserves or reverses the orientation. You can prove this by mimicing the proof of $proposition \ 2.30$ in Hatcher.

So now consider a open ball in $M$ and contract its complement to a point. Then the quotient map will be your required map $f: M\to S^n$ which is degree $1$ by previous discussion.

answered May 20 '16 at 13:51

Anubhav Mukherjee

6,708

I did exactly this 5 minutes ago also mimicing the proof of the local degree formula of Hatcher!:) – noctusraid May 20 '16 at 13:56
I think I worked something out, can I give an answer and have you take a look on it? – noctusraid May 20 '16 at 14:03
Great, thanks!! – noctusraid May 20 '16 at 14:27
what do you think? – noctusraid May 21 '16 at 16:11
I've checked it yesterday only, and It is right...that is why I didnt comment – Anubhav Mukherjee May 21 '16 at 17:52

score 2 · Answer 3 · answered May 20 '16 at 14:21

Thanks to Grumpy Parsnip and Anubhav.K I think I was able to fill in the details. So let $M$ be a closed connected orientable n-manifold.

Let $U_B$ be an open ball in $M$ homeomorphic to some open ball $B^n$ in $\mathbb{R}^n$ and define the quotient $M/\sim$, where for $x \in M$ we have the identification: $x_1 \sim x_2 : \iff x_1,x_2 \in M\setminus U_B$.

Since $U_B$ is homeomorphic to $B^n$ we see that $M / \sim$ is homeomorphic to the sphere $S^n$ and w.l.o.g. the south pole is the point where we have our identification corresponding to the complement of $U_B$. Therefore we have a quotient map $f: M \to S^n$. If we pick a point $y \in S^n$ that is not the south pole, we can choose an open neighborhood $V$ of $y$ in $S^n$ such that $f^{-1}(V)$ is homeomorphic to $V$ and is inside $U_B$. By the local degree formula, that works for local connected n-manifolds, we get that $\deg(f)= \pm 1$. If needed we can compose by a reflection to obtain degree equal 1.

degree 1 map $f : M \to S^n$

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