# Equivariantly split in an isomorphism of $KR$-groups

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.

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.