Using Mayer-Vietoris sequence to show the Möbius band does not embed in $S^2$

Question

I'm studying for an exam and one of the questions is to show that the Möbius band $M$ does not embed into $S^2$. Note that we have $M=[0,1]\times [0,1]/\sim$ where $(0,s)\sim(1,1-s)$

We were given the following hint: Suppose that $f:M\rightarrow S^2$ is an embedding. Let $V=S^2\backslash f(\{(t,1/2):t\in[0,1] \})$ and use the Mayer-Vietoris sequence to compute the homology of $S^2=f(M)\cup V$ to arrive at a contradiction.

Here all homology has coefficients in $\mathbb Z$. I know that $H_n(S^2)=\mathbb Z$ when $n=0,2$ and is trivial otherwise. I also know that $H_n(M)=H_n(S^1)=\mathbb Z$ when $ n=0,1$ and trivial otherwise.

To use the Mayer-Vietoris sequence we will also need to know $H_n(f(M))$, $H_n(V)$ and $H_n(f(M)\cap V)$. But we have $f(M)\cap V= V$ so $H_n(f(M)\cap V)=H_n(V)$.

This is where I am getting confused, I am not sure how to use that I know $H_n(M)$ and that $f$ is an embedding to calculate $H_n(f(M))$. It seems to me that all sorts of things can go wrong under an embedding. Also $H_n(V)$ is $H_n(f(C))$ where $C$ is the center circle of Möbius band. Again it seems that all sorts of things can go wrong under an embedding.

Answer 1

We have $V = S^2 \setminus f(C)$ and thus $f(M) \cap V = f(M \setminus C) \approx M \setminus C$.

Now consider the following part of the Mayer-Vietoris sequence: $$H_1(f(M) \cap V) \stackrel{\phi}{\rightarrow} H_1(f(M)) \oplus H_1(V) \to H_1(S^2) = 0 .$$ This shows that $\phi$ must be onto. To get a contradiction, it is essential to know what $\phi$ looks like: We have $\phi(x) = (i_*(x), j_*(x))$ where $i : f(M) \cap V) \to f(M)$ and $j : f(M) \cap V \to V$ are the inclusions. Since the projection $p : H_1(f(M)) \oplus H_1(V) \to H_1(f(M))$ is onto, we see that $p \circ \phi = i_*$ must be onto. This is equivalent to $k : M \setminus C \hookrightarrow M$ inducing a surjection on $H_1$.

It is well-known (and easy to show) that the boundary circle $S$ of $M$ is a strong deformation retract of $M \setminus C$. Hence the inclusion $l : S \to M \setminus C$ is a homotopy equivalence. Simlarly we have an obvious strong deformation retraction $r : M \to C$. Thus $f = r \circ k \circ l : S \to C$ must induce a surjection on $H_1$. However, it is ea^sy to see that if we identify $S$ and $C$ with $S^1$, then $f$ wraps the circle twice around itself. Thus $f$ corresponds to a map $g : S^1 \to S^1$ of degree $2$. This does not induce a surjection, and we have the desired contradiction.