Inclusion map in Mayer-Vietoris sequence for $\mathbb {RP}^2$ with 3 points removed

Question

I was trying to compute the homology groups of the projective plane with 3 points removed and I was wondering how the map $H_1(j_1,-j_2)$ acts on the first homology group of the intersection.

As open sets $U$ and $V$ I took

$U=(\mathbb{RP}^2\setminus \{P,Q,R\})\setminus\partial\mathbb{RP^2}\sim D^2\setminus\{P,Q,R\}{\xrightarrow{r}}\mathbb S^1\vee \mathbb S^1\vee\mathbb S^1$,

$V$= tubular neighbourhood of the boundary of the projective plane such that $P,Q,R\notin V\sim\mathbb S^1$,

so $U\cap V\sim\mathbb S^1$.

In order to compute the first homology group of $X:=\mathbb {RP^2}\setminus\{P,Q,R\}$ I can use the fact that for path connected topological space $Ab(\pi_1(X,x_0))\cong H_1(X)$ or, since $X$ retracts on the bouquet of 3 circles, we could write $H_1(X)\cong H_1(\mathbb S^1\vee\mathbb S^1\vee\mathbb S^1)\cong \mathbb Z^3$.
But if I try to compute the group with Mayer-Vietoris long sequence I can't figure out the action of the map $H_1(j_1,-j_2):H_1(U\cap V)\to H_1(U)\oplus H_1(V)$ in order to compute $H_1(X)\cong (H_1(U)\oplus H_1(V)) /Ker(H_1(i_1+i_2))$, where $H_1(i_1+i_2):H_1(U)\oplus H_1(V)\to H_1(X)$ and for exactness $Ker(H_1(i_1+i_2))=Im(H_1(j_1,-j_2))$. In particular I should consider the generator of the first homology group of the intersection, let's say $\gamma$, and map it to $ H_1(U)\oplus H_1(V)\cong \mathbb Z^4$.

Thank you so much for your help!

Answer 1

To get a deformation retraction of $X$ onto the bouquet of circles, you should arrange things as in this answer. To use your decomposition to compute homology (without knowing $X\simeq S^1\vee S^1\vee S^1$) and think about the induced maps in the M-V sequence:

$U$ much more obviously retracts onto something homotopy equivalent to $S^1\vee S^1\vee S^1$, this is basically the known disk-minus-points construction. It even deformation retracts onto this thing, so we know $H_\ast(U)$.

$V$ deformation retracts to a circle in the obvious way. $U\cap V$ is also a circle if you make the tubular neighbourhood thin enough but $U\cap V\hookrightarrow V$ is not inducing the identity map as you would expect; because of the antipodal equivalences, our central loop in $U\cap V$ gets mapped under, say, $z\mapsto z^2$ to a doubly wound loop in $V$! $U\cap V\hookrightarrow U$ gives the map $n\mapsto(n,n,n)$ (up to isomorphism); you see this by letting the outermost loop come down onto the triad and notice it winds exactly once around each of the three circles.

So on reduced Mayer-Vietoris we see: $$\cdots\to0\to H_2(X)\to\Bbb Z\overset{(-2,1,1,1)}{\longrightarrow}\Bbb Z^4\to H_1(X)\to0\to0\to0$$From which we conclude $H_2(X)=0$ and $H_1(X)\cong\Bbb Z^3$ can be inferred; map $\Bbb Z^4\to\Bbb Z^3$ via $(a,x,y,z)\mapsto(x+y+a,x-y,z-y)$. This surjects (set $y=0$ to see this) with kernel the span of $(-2,1,1,1)$.