Equivalence of two definitions of $K^{-1}$ in complex $K$-theory In complex $K$-theory, one way I have seen the group $K^{-1}(X)$ defined, for a compact Hausdorff space $X$, is
$$K^{-1}(X):= K^0_c(X\times\mathbb{R}),$$
where the right-hand side refers to compactly supported $K$-theory, which by definition is equal to the reduced $K$-theory of the one-point compactification,  $\widetilde{K}^0((X\times\mathbb{R})^+$).
Meanwhile, the standard way to define the same group seems to be
$$K^{-1}(X):= \widetilde{K}^0(\Sigma X),$$
where $\Sigma X$ is the reduced suspension of $X$.
Question: Is there an easy (perhaps intuitive) way to see that these two definitions coincide?
Thoughts: For example, it seems to me that $(X\times\mathbb{R})^+$ and $\Sigma X$ are not necessarily homotopy equivalent, since one of them involves collapsing the ends of $X\times\mathbb{R}$ to a point, while the other involves doing this as well as collapsing an arc to a point.
 A: You have unreduced $K$-groups $K^0(X)$ which are defined for unbased spaces $X$ and reduced $K$-groups $\tilde K^0(X)$ which are defined for based spaces $X$. The latter should be written more precisely as $\tilde K^0(X,x_0)$, where $x_0 \in X$ is the base-point.
These groups are related by
$$K^0(X) = \tilde K^0(X^+) , \tag{1}$$
where $X^+$ is the topological sum of $X$ and a one-point space $\{\infty\}$. Note that $X^+$ has $\infty$ as base-point.
I do not know how you introduce $K$-theory with compact support, but one way to do it is to define for locally compact unbased $X$
$$K^0_c(X) = \tilde K^0(X^+)$$
where $X^+ = X \cup \{\infty\}$ is the one-point compactification of $X$ with base-point $\infty$ (if $X$ is already compact, then we take $X^+$ as above).
Now you see the problem: Your first definition of $K^{-1}(X)$ is for unbased $X$, your second definition for based $X$. It would be better to write
$$\tilde K^{-1}(X) = \tilde K^0(\Sigma X)$$
in that case.
Anyway, for compact $X$ we should have in analogy to $(1)$
$$K^{-1}(X) = \tilde K^{-1}(X^+) . \tag{2}$$
The LHS is $\tilde K^0$ applied to $((X \times \mathbb R)^+,\infty)$ and the RHS is  $\tilde K^0$ applied to $\Sigma(X^+,\infty)$. But these two spaces are homeomorphic as pointed  spaces.
