3

My reference answer is: https://math.stackexchange.com/a/768843/342943

My problem is:

While writing MW exact sequence

$$\cdots\to H_2(M)\oplus H_2(D^2)\to H_2(\mathbb{R}P^2)\to \underbrace{H_1(S^1)}_{\cong \mathbb Z}\underbrace{\to}_{\times 2} \underbrace{H_1(M)\oplus H_1(D^2)}_{\cong \mathbb Z}\to\cdots$$

This underbraced $\times 2 :\mathbb Z\to \mathbb Z$

where we get that $\times 2$ map in the above sequence from the fact that the inclusion of the intersection of the two spaces (homotopy equivalent to a circle) into the Mobius strip is (up to homotopy) the degree-$2$ covering map, which induces multiplcation by $2$ in first homology.

I don't understand the naturality of this $\times 2$ map since Mayer–Vietoris sequence is given there as(in its general form) $$((j_U)_*,-(j_V)_*):H_n(U\cap V)\to H_n(U)\oplus H_n(V)\\ \text{where}\quad j_U: U\cap V\to U, j_V: U\cap V\to V,\quad \text{inclusions}$$

How can we find this last natural general group homomorphism from this $2-$fold covering. I cannot relate them explicitly.

Arctic Char
  • 16,007
Jale'de jaled
  • 780

1 Answers1

5

Let $\pi : M \to S^1$ be the deformation retraction of the Möbius strip onto the centre circle. If $i$ denotes the inclusion of $S^1$ as the boundary of $M$, then $\pi\circ i : S^1 \to S^1$ is a double covering (compare the green and yellow circles in the image below, taken from here).

$\hspace{34mm}$enter image description here

Therefore the induced map is $(\pi\circ i)_* : H_1(S^1) \xrightarrow{\times 2} H_1(S^1)$ (the degree of a covering map is the number of sheets). As $\pi$ is a homotopy equivalence, $\pi_* : H_1(M) \to H_1(S^1)$ is an isomorphism, so the map $i_* : H_1(S^1) \to H_1(M)$ sends a generator of $H_1(S^1)$ to twice a generator of $H_1(M)$.

$$\require{AMScd} \begin{CD} H_1(S^1) @>{i_*}>> H_1(M) @>{\pi_*}>> H_1(S^1)\\ @VV{\cong}V @VV{\cong}V @VV{\cong}V\\ \mathbb{Z} @>{\times 2}>> \mathbb{Z} @>{\cong}>> \mathbb{Z}. \end{CD}$$

Michael Albanese
  • 99,526