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.)