In section 3, pg 8 of [String theory on Elliptic curves](https://arxiv.org/pdf/1402.4885.pdf) 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



Equivariantly split in an isomorphism of $KR$-groups
Equivariantly split just means [split](https://ncatlab.org/nlab/show/split+monomorphism) 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.


