I've been working computing several K-groups associated to some $C^\*$-algebras involved with my master's thesis, however I've just got stucked finding some generators for $K\_1(C(\mathbb{T})\otimes\mathbb{K})$.


Just to elaborate my question let me explain the analogous problem associated to $K\_0(C(\mathbb{T})\otimes\mathbb{K})$:


It is of course true that 
$$K\_0(C(\mathbb{T})\otimes\mathbb{K})\cong K\_0(C(\mathbb{T}))\cong \mathbb{Z}.$$
Also, it is a well known fact that $[1]$ is a generator of $K\_0(C(\mathbb{T}))$, so one would be tempted to find the generator of $K\_0(C(\mathbb{T})\otimes\mathbb{K})$ using this information. Luckily, we have the next result: 


The map from a $C^\*$-algebra $A$ into $A\otimes\mathbb{K}$ given by $a\mapsto a\otimes e\_{11}$, where $e\_{11}$ is a rank one projection, induces an isomorphism between $K\_0(C(\mathbb{T})$ and $K\_0(C(\mathbb{T})\otimes\mathbb{K})$.


Finally, joining all the pieces it follows that a generator for $K\_0(C(\mathbb{T})\otimes\mathbb{K}$ is $[1\otimes e\_{11}]$.


Since the $K\_1$-functor is also stable and finding generators for $K\_1(C(\mathbb{T}))$ is not too hard, one would be tempted to do the same trick as above, however I have not found any analogous result for the $K\_1$-groups. The main problem (I think) is that the proof (at least the one that I know) of the fact that the $K\_1$-functor is stable is non-constructive, in the sense that the existence of the isomorphism follows from the continuity under direct limits of the $K\_1$-functor. 


With all this said, a generalized question is: Is there any known isomorphism between $K\_1(A)$ and $K\_1(A\otimes\mathbb{K})$? or more precisely (to my goals): Is there any way to find the generators of $K\_1(A\otimes\mathbb{K})$ knowing the generators of $K\_1(A)$?



What are the generator for $K_1(C(\mathbb{T})\otimes\mathbb{K})$?
The morphism $f : A \to A \otimes \mathbb K$ which maps $a$ to $a \otimes e\_{11}$ induce isomorphism $K\_1(f) : K\_1(A) \to K\_1(A \otimes \mathbb K)$ on $K\_1$ groups as well. In fact, we can construct very explicit inverse $KK\_0(A \otimes \mathbb K, A)$ cocycle. Indeed, suppose $\mathbb K$ acts on separable Hilbert space $\mathcal H$, then $A \otimes \mathcal H$ is naturally Fredholm $A \otimes \mathbb K, A$ bimodule, which is naturally cocycle inside $KK\_0(A \otimes \mathbb K, A)$. It's not so hard to see that Kasparov product with the morphism above is an identity.


