I have a problem with understanding the following lines, called the Hierarchy of Compact Surface.
Let $M$ be a compact surface with non-empty boundary. Then $M$ is either closed disc $\Bbb D^2$ or has a non-separating properly embedded arc. And for the latter case, we have a sequence $M=M_0,..., M_l=\Bbb D^2$ of surfaces each obtained from the previous surface by cutting along a non-separating arc, so that $$\text{rank}\big(H_1(M_k;\Bbb Z)\big)-1=\text{rank}\big(H_1(M_{k+1};\Bbb Z)\big)\text{ and }$$$$ l=\text{rank}\big(H_1(M_0;\Bbb Z)\big)=\text{rank}\big(H_1(M;\Bbb Z)\big).$$
I am thinking as follows: $M$ is either $\Sigma_{g,k}:=$ the surface of genus $g$ minus $k$ open disks, so that $H_1(\Sigma_{g,k};\Bbb Z)=\Bbb Z^{2g+k-1}$ or $\Gamma_{n,k}:=\underset{n\text{-times}}{\underbrace{\Bbb RP^2\#\cdots\#\Bbb RP^2}}\bigg\backslash\big(k\text{-open disks}\big)$, so that $H_1(\Gamma_{n,k};\Bbb Z)=\Bbb Z^{n+k-1}$. But $\Sigma_{g,m}$ is nothing but $\Bbb S^2$ with $g$-handles and $m$ open disks removed. Similarly, $\Gamma_{n,k}$ is nothing but sphere with some handles, attaching some Mobius band after removing some open disks, and removing $k$-open disks. For example, consider $M_0=\Sigma_{1,2}$ given below:
Now, I remove the red arc from the handle, so the space obtained is homotopically equivalent to $M_1=\Sigma_{0,4}\cup\alpha$, where $\alpha$ is an arc having exactly two common points with $\Sigma_{0,4}$ such that these two common points lie on two different boundary components of $\Sigma_{0,4}$. Note that we can not remove the blue closed curve as it doesn't change the rank.
$1.$ Am I in the right direction? That is each time I have to remove such red arcs from each handle in any case, orientable or non-orientable?
$2.$ Now, in the case $M$ is non-compact we have $\displaystyle M\simeq \bigvee_{I}\Bbb S^1$ for some index set $I$ having at most countably many points, so $H_1(M;\Bbb Z)=\prod_I\Bbb Z$. Is there any similar notion of hierarchy of non-compact surface, taking inductive limit or inverse limit, etc.?
Thanks in advance, any help will be appreciated.