Calculating degree of a reflection of a single coordinate I am working on the Glen Bredon's ``topology and geometry''. In its proposition 6.10, the degree of a mapping $(x_0, \dots, x_n) \mapsto (-x_0, \dots, x_n)$ from $\mathbb{S}^n \to \mathbb{S}^n$ is introduced. The statement of that proposition is following.
For $\mathbb{S}^n \subseteq \mathbb{R}^{n + 1}$ with coordinates $x_0, \dots, x_n$, let $f: \mathbb{S}^n \to \mathbb{S}^n$ be given by $f(x_0,x_1, \dots, x_n) = (-x_0, x_1, \dots, x_n)$, the reversal of the first coordinate only. Then $\deg(f) = -1$.
Here, $H_*$ denotes any homology theory satisfying the Eilenberg-Steenrod axioms.
As a reminder, the degree of a continuous function from $\mathbb{S}^n$ into $\mathbb{S}^n$ is defined by following.
If $f : \mathbb{S}^n \to \mathbb{S}^n$ is a continuous map, $\deg(f)$ is defined to be the integer such that $f_*(a) = \deg(f) a$ for all $a \in \tilde{H}_n(\mathbb{S}^n ; \mathbb{Z}) \cong \mathbb{Z}$, where $f_*$ denotes the homomorphism of abelian groups induced by $f$ and the functor $H_n$.
My question is following. The proof of the above proposition starts with the case of $n = 0$. Obiovusly, $\mathbb{S}^0 = \{x, y\}$ with $x = 1$, $y = -1$ in $\mathbb{R}^1$. By the additivity of a homology theory, the inclusions $\{x\} \hookrightarrow \mathbb{S}^0$ and $\{y\} \hookrightarrow \mathbb{S}^0$ induce an isomorphism
\begin{equation}
H_0(\{x\}) \bigoplus H_0(\{y\}) \overset{\cong}{\to} H_0(\mathbb{S}^0)
\end{equation}
At this moment, the textbook says that the homomorphism $f_*$ becomes, on the direct sum, the interchange $(a,b) \mapsto (b,a)$, where we indentify the homology of all one-point space via the unique maps between them.
How can I verify the statement about this $f_*$. It is obvious that the map $f$ assigns $y$ to $x$ and $x$ to $y$. However, since I am working with a general homology theory, not a singular homology theory, I cannot exploit the structure of singular complexes, so I have been in trouble with connecting the behaviour of $f$ and $f_*$.
I hope anyone let me know how I proceed. Thank you.
 A: You consider an arbitrary homology theory $H_*$ satsyfing all Eilenberg-Steenrod axioms including the dimension axiom. Thus you should not use the notation $\tilde{H}_n(\mathbb{S}^n ; \mathbb{Z})$ because that suggests that the functors $H_n$ have a second variable (the coefficient group). But such a concept is not contained in the Eilenberg-Steenrod axioms. What you mean is that the coefficient group of $H_*$ is $H_0(*) = \mathbb Z$, where $*$ denotes a one-point space.
We have $S^0 = \{-1,1\} \subset \mathbb R$. The reflection map $f$ simply exchanges the points $\pm 1$. Let $i_\pm : * \to S^0$ denote the map with image $\pm 1$. Note that $f \circ i_\pm = i_\mp$. We know that
$$j :  H_0(*) \oplus  H_0(*) \to H_0(S^0), j(a,b) = (i_-)_*(a) + (i_+)_*(b)$$
is an isomorphism. Let $\phi : H_0(*) \oplus  H_0(*) \to  H_0(*) \oplus  H_0(*), \phi(a,b) = (b,a)$. We claim that the diagram
$\require{AMScd}$
\begin{CD}
H_0(*) \oplus  H_0(*) @>{\phi}>> H_0(*) \oplus  H_0(*) \\
@V{j}VV @V{j}VV \\
H_0(S^0) @>{f_*}>> H_0(S^0) \end{CD}
commutes which answers your question. By the way, to prove this we do not have to know that $j$ is an isomorphism. Moreover, the proof works for any coefficient group.
We have
$$j(\phi(a,b)) = j(b,a) = (i_-)_*(b) + (i_+)_*(a) = (f \circ i_+)_*(b) + (f \circ i_-)_*(a) \\= f_*((i_+)_*(b)) + f_*((i_-)_*(a)) = f_*((i_+)_*(b) + (i_-)_*(a)) = f_*(j(a,b)) .$$
A: There is only a single map from a point to a point, the identity map. Any homology theory satisfying the Eilenberg-Steenrod axioms is a family of functors $H_i: \mathcal{Top}\to \mathcal{Ab}$. And a functor, by definition, takes the identity morphism to the identity morphism.
