Induced map on zeroth homology is zero

I am working through some examples on the homology of mapping torii in Hatcher's Algebraic Topology. One thing that is confusing me is the following:

I don't see why the map on the zeroth homology must be zero. We know that $$\deg g=-1$$ and thus $$g$$ is homotopic to minus the identity. Why then, is the induced map $$\mathbb{1}-g_*=0$$ on zeroth homology?

I'm sure this is simple, but I can't see it myself.

Edit: more generally, how does a degree $$n$$ map act on the zeroth homology? It seems 'obvious' that it should induce the identity map, but why...

• The induced map on zeroeth homology of a map between path-connected spaces is always the identity. Try to prove this, and then generalize to non-path-connected spaces! – user98602 May 16 '16 at 17:16

Recall that the $H_0$ of a path-connected space $X$ is isomorphic to $\mathbb{Z}$ via the augmentation map, which counts the sum of the coefficients of a given (class of a) $0$-chain, i.e., a formal sum of points of $X$. Therefore the generator $1 \in \mathbb{Z}$ corresponds to the class of any point $x \in X$; alternatively, if you pick $x,y \in X$, then they both belong to the same class because their difference is the boundary of a path ($1$-simplex) from one to the other.
Therefore, any continuous $g:X \to X$ induces the identity at the level of $H_0$ because $$1 \leftrightarrow [x] \mapsto [g(x)] \leftrightarrow1,$$
i.e. $1$ is mapped to $1$.