Question about use of set of integers in persistent homology paper

Question

I am unclear what is meant here with the notation $\mathbb{Z}(...)$ and 'extends linearly over $\mathbb{Z}$'. I'm reading this paper 'embedded homology of hypergraphs and applications', and admittedly I don't have a strong formal background here, I'm willing to learn though! Here is the example they give:

A standard n-simplex is denoted as

$$\Delta^n = \{v_0,...,v_n\}.$$

The $(n-1)$-faces of the standard n-simplex $\Delta^n$ are denoted as

$$\Delta_i^{n-1} = \{ v_0,...,\hat{v}_i,...,v_n \}, 0 \leq i \leq n$$

We have the face maps $d_i$ sending $\Delta^n$ to $\Delta_i^{n-1}$, for $0 \leq i \leq n$. And we have the boundary maps

$$\partial_n:\mathbb{Z} \to \mathbb{Z}(\Delta_0^{n-1},...,\Delta_n^{n-1})$$

given by $\partial_n = \sum_{i=0}^n(-1)^id_i$, which extends linearly over $\mathbb{Z}$. The power set of $\Delta^n$ is denoted as $\Delta[n]$, called the standard simplicial complex spanned by $n+1$ vertices.

Answer 1

For any set $S$, the free abelian group on $S$, denoted as $\mathbb Z(S)$, is defined to be the set of all finite sums$^\color{blue}*$ $x = n_1s_1 + \cdots + n_ks_k$, where $k \in \mathbb Z^+$, $n_i \in \mathbb Z$ and $s_i \in S$, with the same $+$ as operation. It has the property that for any abelian group $G$ and any function $f : S \to G$ there exists a unique group homomorphism $\widetilde{f\,}\! : \mathbb Z(S) \to G$ that extends $f$ (i.e. $\widetilde{f\,}\!(s) = f(s)$ for all $s \in S$), namely, $$\widetilde{f\,}\!(n_1s_1 + \cdots + n_ks_k) = n_1f(s_1) + \cdots + n_kf(s_k).$$ We say that $\widetilde{f\,}$ is obtained from $f$ by extending it linearly over $\mathbb Z$, and sometimes $\widetilde{f\,}$ is also denoted by the same symbol $f$.

In your case, the function $\partial_n : \Delta^n \to \mathbb Z(\Delta_0^{n-1},\dots,\Delta_n^{n-1}) := \mathbb Z(\Delta_0^{n-1} \cup \cdots \cup \Delta_n^{n-1})$ given by $v \mapsto \sum_{i=0}^n (-1)^id_i(v)$ extends linearly over $\mathbb Z$ to a homomorphism $\partial_n : \mathbb Z(\Delta^n) \to \mathbb Z(\Delta_0^{n-1},\dots,\Delta_n^{n-1})$.

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$^\color{blue}*$Formally, $\mathbb Z(S)$ consists of functions $x : S \to \mathbb Z$ such that $\{s \in S : x(s) \neq 0\}$ is finite, with pointwise addition. Note then that, for each $s \in S$, the function (which I will denote by the same letter $s$) that sends $s$ to $1$ and everything else to $0$ is an element of $\mathbb Z(S)$; and so, for each $x \in \mathbb Z(S)$ we have that $x$ is a finite linear combination of "elements" in $S$: $$x = \sum_{s \in S}x(s)s.$$

Answer 2

aziff00's answer is neat and clear from the formal point of view. Let me add something on motivations and intuition.

First note that in the formula for $\partial_n$ the only integers appearing in the formula are $\pm 1$. This can be thought of as orienting the faces in a coherent manner. The easiest case being that the boundary of an interval (1-simplex) $PQ$ is just $Q-P$.

Passing from $\pm$ sign to integers combinations of simplexes means just assigning a pair of multiplicity $|n|$ and sign $\mathrm{sgn}(n)$ to each simplex and extending the face maps to be $\mathbb Z$-linear just means that faces of a simplex with multiplicity say $4$ all will have multiplicity $4$; like if you take $4$ copies of the same triangle then in its boundary each side will have multiplicity $4$.

So the formal integer combinations just means that you can consider collections of simplices with multiplicity and orientations. The boundary map preserves multiplicities and orientation.