I am reading an alternative way of calculating $H_1(\mathbb{R}P^2)$ not through the use of delta complexes and they have used the following fact:
$H_1(M, \delta M) \cong H_1(\mathbb{R}P^2,D)$ where $M$ is the mobius strip and $D$ is the disk
The prior part of this proof uses excision to calculate $H_1(M,\delta M) \cong \mathbb{Z_2}$ which i understand but my question is how does excision imply the statement above.
I know that $\mathbb{R}P^2$ is composed of $M$ and $D$ along the boundary $S^1$ of M, and that $S^1 \subset M \subset \mathbb{R}P^2$ but id really like an explanation of why this holds before i continue
thanks for the help in advance
Okay we want $U\subset D$ for excision, eg $\frac 12 \stackrel \circ D$. As you said: $RP^2-U$ is a Mobius Band. You have two easy ways to write down a homotopy equivalence of the pairs $(RP^2-U,D-U)$ and $(M,\partial M)$. Then excision tells you $$ H_1(RP^2)=H_1(RP^2,D)=H_1(RP^2-U,D-U)=H_1(M,\partial M) $$