What are the generator for $K_1(C(\mathbb{T})\otimes\mathbb{K})$?

Question

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 $K_1(A)$?

Answer 1

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.