In Higson and Roe's Analytic K-homology, for a unital $C\*$-algebra $A$, the definitions of K-theory and K-homology have quite a similar flavor. 


Roughly, the group $K\_0(A)$ is given by the Grothendieck group of homotopy classes of matrix algebra projections over $A$.


On the other hand, $K^0(A)$ is the Grothendieck group of homotopy classes of even Fredholm modules over $A$ (or more correctly unitary classes of Fredholm modules). 


Now in contrast to the $K\_0(A)$ case, every element of $K^0(A)$ contains a representative Fredholm module. (For the inverse of the class of $(H,\rho,F)$, just take the class of $(H^{\text{op}},\rho,-F)$, where op denotes the opposite grading. 


Does this means that taking the Grothendieck is not strictly necessarily to define $K^0(A)$? Could one just as well define it as the monoid of homotopy classes of Fredholm modules, and then prove that it was a group?


If this is true, then is there any deeper philosophical reason why $K\_0$ requires us to introduce inverses, while $K^0$ does not?



Comparing the definitions of $K$-theory and $K$-homology for $C^*$-algebras
KK-theory provides one way to eliminate the asymmetry in the definitions of K-theory and K-homology. A cycle in $KK(A, B)$ is a triple $(H, \rho, F)$ where $H$ is a (adjectives) Hilbert $B$-module, $\rho$ is a (adjectives) representation of $A$ on $H$, and $F$ is a (adjectives) bounded operator on $H$ such that $[F, \rho(a)]$, $(F^2 - 1)\rho(a)$, and $(F - F^\*)\rho(a)$ are $B$-compact for all $a \in A$. The relations are analogous to the relations in K-homology, and in particular KK-cycles form an abelian group without the need to invert anything.


If $B = \mathbb{C}$ then $H$ is an ordinary Hilbert space and the KK-cycles are exactly the same thing as Fredholm modules over $A$. It is a fact that $KK(\mathbb{C}, B)$ is isomorphic to the K-theory of $B$ for all (adjectives) C\*-algebras $B$, so this gives a model of K-theory of the sort you're looking for.


