# Clarification of “death event” in persistent homology

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:

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

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

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

The main idea is the following. Since $$\alpha$$ was born at the $$i$$th level, the bullet point $$f_r^{i,j-1}(\alpha)\notin \operatorname{im}(f_r^{i-1,j-1})$$ implies that $$\alpha$$ is still alive at index $$j-1$$. Roughly speaking, this is because if we suppose for a contradiction that $$f_r^{i,j-1}(\alpha)\in \operatorname{im}(f_r^{i-1,j-1})$$, then this would instead imply that $$\alpha$$ were born earlier (at or before index $$i-1$$). Additionally, the bullet point $$f_r^{i,j}(\alpha)\in \operatorname{im}(f_r^{i-1,j})$$ implies that $$\alpha$$ is dead by index $$j$$. Roughly speaking, this is because if we suppose for a contradiction that $$f_r^{i,j}(\alpha)\notin \operatorname{im}(f_r^{i-1,j})$$, 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 $$j$$.
• 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? – gian Oct 10 '17 at 15:28
• 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. – Henry Adams Oct 13 '17 at 19:19