I apologize in advance about the pictures...I am not sure how to properly crop/rotate them.
Suppose $K: S^1\to S^3$ is a map whose image is the trefoil knot. Let $X=S^3\setminus K$ (identifying $K$ with its image in $S^3$). We know (from class) that $\pi_1 X=\langle x,y \;| \;xyx=yxy\rangle$. Since $K\subset S^3$ is a knot, it bounds an orientable surface (call it $F$). $F$ is like 2 disks that are joined by 3 twisted bands. Here is a picture of what $F$ looks like (up to homotopy):
We are then asked to find use Mayer-Vietoris sequence to find $H_1(Y)$ if $Y$ is the two fold cover of $X$ obtained from the surjective map $\phi: \pi_1 X\to \mathbb{Z}/2\mathbb{Z}$.
Here is what I have tried: Using the group presentation for $\pi_1 X$ above and $\mathbb{Z}/2\mathbb{Z}\cong \{1,-1\}$, I first want to understand $\phi$. So it suffices to see where $\phi$ sends the generators $x$ and $y$. If $\phi(x)=1$, then $\phi(y)=\phi(y^2)=\phi(y)^2$ and so $\phi(y)=1$ (but then $\phi$ not surjective). Similarly, we can't have $\phi(y)=1$. So $\phi(x)=\phi(y)=-1$. Now I want to understand $Y$. I think $Y$ is supposed to look something like this:
So its like two copies of $X$, with the two "end" copies of $F$ identified. This sort of makes sense intuitively since $x$ and $y$ each represent a loop in $S^3$ that wraps around $K$ (and so it passes through $F$ once). The fact that $\phi(x)=\phi(y)=-1$ inticates to me that going around $x$ (or $y$) in $X$ should result in "switching" copies of $X$ in $Y$. My first question is: How exactly do I make this precise?
My other question, is how exactly do I use Mayer-Vietoris sequence to find $H_1(Y)$? I know that I should choose appropriate $A,B$ such that $A\cup B=Y$ and then use the exact sequence $$ H_1(A\cap B)\to H_1(A)\oplus H_1(B)\to H_1(Y)\to H_0(A\cap B)\to H_0(A)\oplus H_1(B) $$ to conclude what I want about $H_1(Y)$. For instance, maybe I can choose $A,B$ so that $H_0(A\cap B)=H_1(A\cap B)=0$ so that I get an isomorphism $$ H_1(A)\oplus H_1(B)\cong H_1(Y). $$ For instance, should I choose $A,B$ as drawn below? ($A$ shaded in blue, $B$ in red).
If I make this choice, then $A\cap B=\emptyset$. So it would suffice to calculate $H_1(A), H_1(B)$. EDIT: One cannot make this choice as you need the INTERIORS of $A$ and $B$ to cover $Y$. But if we make the following choice for $A$ (pink) and $B$ (blue)
then I can use this in Mayer-Vietoris to conclude that $H_1(Y)\cong \mathbb{Z}$ which seems to agree with the comment below that $H_1(Y)\cong ker(\phi)_{ab}$
END EDIT.
Does this seem like the right thing to do?
I would really appreciate a little bit of guidance in this problem... Our school has been closed due to COVID-19 so help on HW has been a little hard to come by. Thank you for your time!