Why is induced map on zero homology the identity and not negative the identity? Suppose we have a simplicial map $f$ on a path connected simplicial complex $X$. The answer here: Induced map on zeroth homology is zero claims that the induced map on the $0$-homology given by $f_*: H_0(X) \to H_0(X)$ is the identity map. As $X$ is path connected I see why $H_0(X)$ is $\mathbb{Z}$ and why the equivalence class of any vertex of $X$ is a generator of $H_0(X)$ corresponding to $1$ in $\mathbb{Z}$. 
The linked answer then continues to say that $f_*$ sends generators of $H_0(X)$ to generators of $H_0(X)$ which I understand. It then claims that for any vertex $x$ we have $1 \leftrightarrow [x]\to[f(x)] \leftrightarrow 1$ so the map must be the identity. However I don't understand why we can't have $[f(x)] \leftrightarrow -1$ as $-1$ is also a generator of $\mathbb{Z}$. Can someone please clear up this confusion for me? Thanks!
 A: Let $X$ be an abstract simplicial complex and $\lvert X \rvert$ its geometric realization which is a topological space representable as the union of geometric simplices. It is well-known that there exist natural isomorphisms $\phi_n^X : H_n(X) \to H_n(\lvert X \rvert)$ between the simplicial homology groups $H_n(X)$ of $X$ and the singular homology groups $H_n(\lvert X \rvert)$ of $\lvert X \rvert$. In particular, if $f : X\to Y$ is a simplicial map, then $\lvert f \rvert_* \circ \phi_n^X = \phi_n^Y \circ f_*$.
Using this, Qiaochu Yuan's comment answers your question. However, we can also see this without changing to topological spaces.
The simplicial chain complex of $X$ is based on oriented simplices. An orientation of an $n$-simplex $s^n = \{ v_0,\dots,v_n \}$ is an equivalence class $[v_{\pi(0)},\dots,v_{\pi(n)}]$, where $\pi$ ranges over all permutations of $\{ 0,\dots,n \}$. For $n > 0$ there exist two orientations, for $n = 0$ only one. The simplicial chain map $C_*(f) : C_*(X) \to C_*(Y)$ is given on the simplicial generators $\sigma^n = [v_0,\dots,v_n]$ of the free abelian group $C_n(X)$ by
$$C_n(f)([v_0,\dots,v_n]) = [f(v_0),\dots,f(v_n)].$$
In dimension $0$ we simply have
$$(*) \phantom{xx} C_0(f)(v) = f(v)$$
with the vertices $v$. 
Now you consider a simplicial map $f : X \to X$ on a path connected $X$. In my understanding of your question you know that all vertices $v$ of $X$ (i.e. all simplicial generators of $C_0(X)$) are in the same homology class $g \in H_0(X)$ and that $H_0(X)$ is free abelian with $g$ as a generator. Let us call $g$ the canonical simplicial generator of $H_0(X)$. This provides a canonical isomorphism $H_0(X) \approx \mathbb{Z}$ (mapping $g$ to $1$). $H_0(X)$ of course has an alternative generator: This is $-g$.
But now it obvious from $(*)$ that $f_*(g) = g$ since both $v$ and $f(v)$ are vertices and therefore in the same homology class.
