Calculating homology of a Klein bottle (Using only axioms)

Calculating homology of a Klein bottle (Using only axioms)

so let K be the Klein Bottle. I am perfectly aware how to calculate it for
singular homology, using some properties of chains, and an explicit
description of the differential. This is not my problem.
Let us take a (unreduced) homology theory and suppose that we have that
$$h_n(S^1,\emptyset)=\mathbb{Z}$$ if $n=0,1$ and otherwise $0$. I want to
calculate the homology of the Klein Bottle using only this fact, and that
the functors $h_n$ satisfy the Eilenberg–Steenrod axioms. On page 109 of
Switzer's book on Algebraic Topology, he shows how this can be done. The
idea is to cut the klein bottle $K$ into two cylinders $A,B$ and that $A
\cap B \cong S^1 \amalg S^1$ and use Mayer-Vietoris: $$\cdots \rightarrow
h_n(A \cap B, \emptyset) \xrightarrow{\alpha} h_n(A,\emptyset) \oplus
h_n(B, \emptyset) \xrightarrow{\beta} h_n(K, \emptyset)
\xrightarrow{\Delta'} \cdots$$ and that $h_n(A\cap B, \emptyset) =
\mathbb{Z} \oplus \mathbb{Z}$, $h_n(A,\emptyset) =\mathbb{Z}$,
$h_n(B,\emptyset) = \mathbb{Z}$. One can show that the map $\beta$ is
represented by a matrix $$\pmatrix{1 & 1 \\ 1 & v'_\ast}$$ where $v':S^1
\rightarrow S^1$ is the map reversing the orientation. Switzer then states
that, $v_\ast:H^1(S^1,\emptyset) \rightarrow H^1(S^1,\emptyset)$ is $-1$
and $H^0(S^1,\emptyset) \rightarrow H^0(S^1,\emptyset)$ is $1$. Now here
is my question (finally): How can we determine the sign of the map? It is
easy to see that the map is either $\pm 1$, but the sign is more
mysterious...