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

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.

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.
• @EpsilonDelta No, $H_1$ is covariant functor, ad induced maps are written as $f_*$. Jan 23, 2020 at 20:02