For a sequence of nested complexes $K_1\subset K_2 \subset K_3$, I have calculated the first homology groups at each level,
$$ H_1(K_1)=\left<a,b\right>\cong \mathbb{Z}^2\\ H_1(K_2)=\left<a,b,c\right>\cong \mathbb{Z}^3\\ H_1(K_3)=\left<a,b,c\right>/\left<a+b+c\right>\cong\left<a,b\right>\cong\mathbb{Z}^2\\ $$ where $a$, $b$, and $c$ are 1-simplexes and the homology groups are constructed with integer coefficients from these generating elements.
In this scenario, how do I go about drawing the barcode for $H_1$? I know the rank of $H_1$ should match the number of generating elements, so it should be $2,3,2$ in this case.
For the first two groups ($H_1(K_1)$ and $H_1(K_2)$) I could imagine taking one bar for the element $a$ and one bar for $b$, that are both present at all stages. However, the quotient $H_1(K_3)$ is also isomorphic to $\left<b,c\right>$, so using this strategy can lead to at least two very different barcodes (see image below)?
This represents whether $c$ is used as a generating element in $H_1(K_3)$ or not. The rank (number of bars) is correct for both barcodes. Anyway, it seems I am wrong on how to construct barcodes since I expected a single correct answer. How should one go about this?
When looking at filtered modules $\mathbb{V} = \{V_t\}$, it's useful to consider the ranks over an interval in addition to the ranks just at single time points, i.e., look at $$\beta^{s,t} = \operatorname{rank}(V_s \to V_t)$$ not just when $s = t$, but also for $s < t$.
In your example, consider the homomorphism $H_1(K_1) \to H_1(K_3)$, which is an isomorphism and has rank 2. So the barcode should contain two full bars spanning from beginning to end. This is consistent with your first answer but not the second.
In general, this is encapsulated in the so-called elder rule: when two classes become homologous, the older one survives and the death should be paired with the class that was born more recently.