Does system of cutting arcs generate $H_1(M,\partial M)$?

Let $$M$$ be a connected, orientable manifold with boundary. Is it true that a set of properly immersed arcs that cut $$M$$ into a ball generates $$H_1(M,\partial M, \mathbb{Z})$$?

If so, are those generators linearly indipendent if the set of arc is minimal?

You must mean $$M$$ a surface, in which case yes (think about the polygonal representation of the closed surface and delete some well-chosen balls). For higher-dimensional $$M$$, deleting arcs will either (1) in 3D, make the fundamanetal group more complicated, or (2) in dimensions 4 and up, not change the fundamental group at all.
In particular you cannot possibly make $$M$$ simply connected by deleting arcs except in dimension 2.
• I wonder if in dimension $n$, one can find a system of some $(n-1)$-discs that cut the manifold into a disc. – giulio Jan 29 at 14:12
• Ah, no, I don't think so. You want the discs to have boundary contained in the boundary, right? Then deleting a disc can only increase the number of boundary components. Dimension 2 is special because we use arcs and $\partial I = 0 \cup 1$ is disconnected. – kkot jon Jan 29 at 14:24
• Correct. But If we don't ask for the disc to have boundary in the boundary of the manifold I think we could do that. Take a handle decomposition and for each $k$-handle $D^k\times D^{n-k}$ remove the disc $D^{k-1}\times D^{n-k}$. The cut handle retracts to the boundary and we finally get a $0$-handle (assuming that the decomposition has only one 0-handle, but if the manifold is connected that's fine). Now, does it make sense to ask if these discs generate $H_{n-1}(M,\partial M)$? – giulio Jan 29 at 14:39