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# How do you intepret "kill a cohomology class" intuitively for attaching an n-cell?

I ran into this problem when studying Morse theory. My professor referred to the torus and riemann surface of genus 2 as an example. And after some manipulating on the long exact sequence of cohomology groups he came to the conclusion that attaching an n-cell(passing a critical point of index n) would either kill a class of m-1 or give birth to a class of m. Denoting $\mathcal{M}^{\pm}=f^{-1}(-\infty, p\pm\epsilon)$, and the index of critical point p is m, here is the long exact sequence of cohomology group:

$0\rightarrow\cdots\rightarrow H^{\ast-1}(\mathcal{M}^-)\rightarrow H^{\ast}(\mathcal{M}^+,\mathcal{M}^-)\rightarrow H^{\ast}(\mathcal{M}^+)\rightarrow H^{\ast}(\mathcal{M}^-)\rightarrow H^{\ast+1}(\mathcal{M}^+,\mathcal{M}^-)\rightarrow\cdots\rightarrow0$

My first question is, is it true that the map $H^{\ast}(\mathcal{M}^+,\mathcal{M}^-)\rightarrow H^{\ast}(\mathcal{M}^+)$ is either injective or the 0 map? Why or why not? And also why $H^{\ast}(\mathcal{M}^+,\mathcal{M}^-)$ is $\mathbb{Q}$ for $\ast=m$ and 0 otherwise?

Secondly, if this is true, we will arrive at the desired results. Either

$0\rightarrow H^{m-1}(\mathcal{M}^+)\rightarrow H^{m-1}(\mathcal{M}^-)\rightarrow0\rightarrow \mathbb{Q}\rightarrow H^{m}(\mathcal{M}^+)\rightarrow H^{m}(\mathcal{M}^-)\rightarrow0$

or

$0\rightarrow H^{m-1}(\mathcal{M}^+)\rightarrow H^{m-1}(\mathcal{M}^-)\rightarrow\mathbb{Q}\rightarrow0$

Thus the "either kill or give birth" statement is obtained. But what I cannot understand is how to view this inuitively, say, how can you tell whether attaching a cell will kill a class or give a new class from a geometric point of view? My professor tried to explain this using torus but I still fail to see it. I just didn't know how he managed to "see" it. I would really appreciate it if someone could kindly tell me how to see it. Thanks.