Mapping torus of Klein bottle, from discussion in Hatcher p. 152. At the very bottom of page 151 to the top of 152 in Algebraic Topology by Hatcher, it says

In the case of the mapping torus of a reflection $g:S^1\to S^1$, with $Z$ a Klein bottle, the exact sequence is
  $$
0\to H_2(Z)\to H_1(S^1)\xrightarrow{1-g_*}H_1(S^1)\to H_1(Z)\to H_0(S^1)\xrightarrow{1-g_*}H_0(S^1)
$$

Since $H_1(S^1)\cong H_0(S^1)\cong \mathbb{Z}$, it says $H_1(S^1)\to H_1(S^1)$ is just multiplication by $2$, and $H_0(S^1)\to H_0(S^1)$ is multiplication by $0$.
I thought that $g$ has degree $-1$ as a reflection, so if $[\alpha]$ is a generator of $H_1(S^1)$ (or $H_0(S^1)$), isn't
$$
(1-g_*)([\alpha])=[\alpha]-(-1)[\alpha]=2[\alpha]?
$$
Why isn't $1-g_*$ multiplication by $2$ in both cases? Thanks.

$g$ does have degree $-1$. Degree only describes the action of a map on top homology, i.e. $H_1$ in this case. Every map between connected spaces acts as the identity on zeroth homology: for any chain $\alpha=\sum a_i \sigma_i, g_*(\alpha)=\sum a_ig(\sigma_i)$. In particular, for a 0-chain $x$, $g_*(x)=g(x)$, which is a generator of $H_0$. The confusion here is between the action of $g$ on the coefficients of a chain and the possibility that $g$ may change the sign of some cycle: the thing to remember is simply that $g$ cannot ever do anything to the coefficients. Thus since a 0-manifold has no internal structure to change, $g$ cannot ever act interestingly on $H_0$. (The caveat is for spaces which aren't connected. Then a map acts on $H_0$ as a matrix all of $1$s and $0$s whose $1$s tell you which connected components map where.)
