Definition of odd topological K-theory using circles I wanted to check whether the following characterization of odd complex topological $K$-theory is correct.
Let $X$ be a compact Hausdorff space. Then $K^{-1}(X)$ can be defined as $\tilde{K}^0((X\times\mathbb{R})^+)$, where $^+$ means one-point compactification.
Let $S^1\subseteq\mathbb{C}$ be the unit circle. Since $(X\times\mathbb{R})^+$ is homeomorphic to $(X\times S^1)/(X\times\{1\})$,
Question: Can we say that an element of $\tilde{K}^0((X\times\mathbb{R})^+)$ is a difference of vector bundles $E$ and $F$ over $X\times S^1$ such that $E$ and $F$ are trivial over $X\times\{1\}$ and the virtual vector bundle $E-F$ has virtual dimension $0$ over $X\times\{1\}$?

I am not sure what is the virtual dimension, but think your point may be essentially correct.
In general, we have the isomorphism
$$K^0(X\times S^1)\cong K^1(X)\oplus K^0(X)$$
and hence we can define $K^1(X)=\ker(K^0(X\times S^1)\rightarrow K^0(X))$. It is exactly the first definition in the origional paper Vector bundles and homogeneous spaces of Atiyah--Hirzebruch for topological $K$-theory. Here the map
$$K^0(X\times S^1)\rightarrow K^0(X)$$
is given by the pullback along $i:X\times\{1\}\hookrightarrow X\times S^1$. By construction, we know that pullback along a subset induces exactly the restriction i.e. $i^*E=E|_{X\times\{1\}}$ for any vector bundle $E$. In the ring $K^0(X\times S^1)$, every element is the difference $[E]-[F]$, and it is mapped to $[f^*E]-[f^*F]$ in $K^0(X)$. The image is trivial if and only if $[f^*E]=[f^*F]$ or equivalently that $E|_{X\times\{1\}}$ and $F|_{X\times\{1\}}$ are stably equivalent.
