# Two fold map while calculating Mayer–Vietoris sequence of $\mathbb RP^2$.

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.

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