Equivariantly split in an isomorphism of $KR$-groups

Question

In section 3, pg 8 of String theory on Elliptic curves they claim that for $X$ compact with $x_0$ an involution fixed point, the map $i:x_o\rightarrow X$ is equivariant and equivariantly split, which according to them implies

$KR^{j}(X) \approx KR^{j}(X - \{x_0 \}) \oplus KO^{j}(x_0)$

I don't understand what equivariantly split means here relative to the functor $KR^j( - )$ other than you can simply do this separation nor why this is so.

1) I would appreciate if someone could put in more steps into the isomorphism $KR^{j}(X) \approx KR^{j}(X - \{x_0 \}) \oplus KO^{j}(x_0)$

2) Is there a reference were they explain in more detail what is meant by equivariantly split? I haven't seen any mention of the term the way is being used here for any kind of equivariant cohomology theory

Note that it seems to be $KR$-theory with compact supports but I don't know if this is essential for the above argument.

Thank you for your time

Answer 1

Equivariantly split just means split in the category of spaces with a $C_2$-action, i.e. that there is an equivariant map $r:X→\{x_0\}$ such that $r\circ i$ is equivalent to the identity. This map is usually called a retraction of $i$. Of course the splitting in this case is trivial to construct.

Since $KR^j(-)$ is a functor from the category of spaces with a $C_2$-action to abelian groups, you have that the map $KO^j(*)=KR^j(\{x_0\})\leftarrow KR^j(X)$ is split too, but in the category of abelian groups this means precisely that if $KR^j(X,x_0):=\ker i^*$ we have $KR^j(X)\cong KR^j(X,x_0)\oplus KO^j(*)$.

To show that $KR^j(X,x_0)\cong KR^j(X\smallsetminus\{x_0\})$ in favorable cases, we are going to assume that $x_0$ has a filter of $C_2$-contractible neighborhoods $\{U_i\}$ (this is the case, e.g., if $X$ is a $C_2$-manifold). Then $KR$-cohomology with compact supports is, by definition, $$KR^j(X\smallsetminus\{x_0\})=\lim_i KR^j(X\smallsetminus U_i) ,.$$ Since by excision the map $KR^j(X\smallsetminus U_i)\cong KR^j(X,x_0)$ is an equivalence, we are done.