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Before I ask my question let me clarify some notation: $f^{i,j}_r$, where $i < j$, refers to the inclusion map $f: H_r(X_i) \hookrightarrow H_r(X_j)$. $X_i$ and $X_j$ are subcomplexes of a filtered complex $X$, where $X_i \subseteq X_j$.

I need clarification/explanation of the following definition. I was able to understand birth events easily, but I'm having some trouble with death events. Why do the two conditions below imply the end of a certain feature?

Suppose an $r$-class $\alpha$ is born at the $i$th level. Then $\alpha$ is said to die at level $j$ iff:

  • $f^{i,j}_r(\alpha) \in im(f^{i-1, j}_r)$
  • $f^{i,j-1}_r(\alpha) \notin im(f^{i-1, j-1}_r)$

A death event then refers to the death of all homology classes in the coset $\alpha + im(f^{i-1,i}_r)$.

The bounty is for whoever answers my question in the comments below.

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This definition of a "death event" is quite closely linked to the related definition of a "birth event", so it may be helpful to post that definition in the statement of your question as well. Also, which reference is this from?

The main idea is the following. Since $\alpha$ was born at the $i$th level, the bullet point $f_r^{i,j-1}(\alpha)\notin im(f_r^{i-1,j-1})$ implies that $\alpha$ is still alive at index $j-1$. Roughly speaking, this is because if we suppose for a contradiction that $f_r^{i,j-1}(\alpha)\in im(f_r^{i-1,j-1})$, then this would instead imply that $\alpha$ were born earlier (at or before index $i-1$). Additionally, the bullet point $f_r^{i,j}(\alpha)\in im(f_r^{i-1,j})$ implies that $\alpha$ is dead by index $j$. Roughly speaking, this is because if we suppose for a contradiction that $f_r^{i,j}(\alpha)\notin im(f_r^{i-1,j})$, then this would instead imply that $\alpha$ were still alive at index $j$. Together these two bullet points imply that $\alpha$ dies exactly at index $j$.

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  • $\begingroup$ Here is the resource: paulbendich.com/pubs/analyz.pdf $\endgroup$ – gian Oct 10 '17 at 13:26
  • $\begingroup$ I think I understand. But what's the intuition behind $\alpha$ and its images having to remain outside the image of a function from $i-1$? Why must $\alpha$ remain "outside" in order to persist? $\endgroup$ – gian Oct 10 '17 at 15:28
  • $\begingroup$ Roughly speaking, if $\alpha$ and its images were inside the image of a function from $i-1$, then $\alpha$ would not have been born at index $i$, but instead at index $i-1$ or earlier. $\endgroup$ – Henry Adams Oct 13 '17 at 19:19
  • $\begingroup$ Ah, I see. So how does one normally keep track of these death events in practice? $\endgroup$ – gian Oct 13 '17 at 20:37
  • $\begingroup$ In practice one computes all of this using linear algebra! See for example the paper "Computing Persistent Homology" by Zomorodian and Carlsson: link.springer.com/article/10.1007/s00454-004-1146-y $\endgroup$ – Henry Adams Oct 27 '17 at 22:37

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