equivalence in the sense of definition \[defgeomk\] is clear.

For the converse, suppose that $(W,f,  \sigma)$ is a bordism from the cycle $(M_0, f_0, \sigma_0)$ to $(M_1, f_1,  \sigma_1)$. Choose a normal bundle $NW$ for $TW$ with respect to $f$ in a way that $NW\mid_{\partial W}$ is a normal bundle for the boundary. Then $NW\bigstar W$ gives a normal bordism between ${\operatorname{Spin}}(c)$-modifications along normal bundles for the boundary components.

The vector bundle modification procedure also implies normal bordism. Let $E \to  M$ be a real, ${\operatorname{Spin}}(c)$, $G$-equivariant vector bundle. Let $F\to S(E\oplus {{\mathbb R}})$ be a ${\operatorname{Spin}}(c)$, $G$-equivariant real vector bundle, and let $N$ be a normal bundle with respect to the map $S(E\oplus  {{\mathbb R}})\to M \to X$ .

Consider the map $s:M\to  F$ given as the composition of the $1$-section $s: M\to S(E\oplus  {{\mathbb R}})$, and the $0$-section $S(E\oplus {{\mathbb R}})\to F$.

Now, consider the bundle $s^*(N)\oplus  F \mid M $, and notice that it is a normal bundle for $M$ with respect to the map $f$.

After lemma \[lemmanormalbordism\], these total spaces are bordant, proper $G$- ${\operatorname{Spin}}(c)$-manifolds. The bordism determines a bordism of geometric cycles.

We now prove the usual homological properties of twisted equivariant $K$-homology for 