$K_0(SS\mathbb{C})$ from the definition

Question

For any $C^\ast$-algebra $A$ its suspension is the $C^\ast$-algebra of all continuous functions $f$ from the unit sphere $S^1$ to $A$ with $f(1)=0$. For the K-theory of $C^\ast$-algebras on has the Bott periodicity $K_0(SSA) \cong K_0(A)$.

How does this general fact work out if one chooses $A = \mathbb{C}$? On the one hand, we know theoretically that $ K_0(SS \mathbb{C}) = K_0(\mathbb{C}) = \mathbb{Z}$. On the other hand, the basic definition of $K_0(SS\mathbb{C})$ misleadingly seems to suggest that $K_0(SS\mathbb{C})$ should actually be zero. Here is why:

1. An element of $SS \mathbb{C}$ is a continuous function from the unit square to $\mathbb{C}$ which vanishes on the sides of the square.
2. By adjoining a unit to $SS \mathbb{C}$ we obtain $(SS \mathbb{C})^+$ which is the algebra of all continuous functions from the unit square to $\mathbb{C}$ with constant value on the sides.
3. For every $n \in \mathbb{N}$, $M_n((SS \mathbb{C})^+)$ is the algebra of continuous functions from the unit square to $\mathbb{C}^{n \times n}$ which assume constant values on the sides.
4. Let $M_{\infty}((SS\mathbb{C})^+) := \bigcup_{n \in \mathbb{N}} M_n((SS\mathbb{C})^+)$ and write $p \sim q$ for projections $p,q \in M_\infty((SS\mathbb{C})^+)$ if there is a partial isometry $v \in M_\infty((SS\mathbb{C})^+)$ with $p = v v^\ast, q = v^\ast v$. As $p$ is a function from the unit square to the projections in some $\mathbb{C}^{n \times n}$, and the rank of a projection is continuous, it seems clear that $p \sim q$ holds iff $p$ and $q$ have the same rank. Thus, $K_{00}((SS\mathbb{C})^+) = \mathbb{Z}$.
5. Let $\pi: (SS\mathbb{C})^+ \to \mathbb{C}$ map a function $f \in (SS\mathbb{C})^+$ to its value on the sides of the unit square. Then $\pi_\ast: K_{00}((SS\mathbb{C})^+) \to K_0(\mathbb{C}) = \mathbb{Z}$ maps the equivalence class of some projection $p$ to the rank of $p$. In particular, the kernel of $\pi_\ast$ contains only $[0]$. Hence, $K_0(SS\mathbb{C}) = \mathrm{ker}(\pi_\ast) = 0$.

What is wrong with this argument?

Answer 1

As $p$ is a function from the unit square to the projections in some $\mathbb{C}^{n \times n}$, and the rank of a projection is continuous, it seems clear that $p \sim q$ holds iff $p$ and $q$ have the same rank.

This step is wrong (as usual, "it seems clear" is a major warning - you haven't actually proven it). You've forgotten the condition that functions have to take constant values on the sides. The unit square with sides identified to a point is the $2$-sphere $S^2$ (the double suspension of a point) and this business with projections amounts to classifying vector bundles on the $2$-sphere; in this step you're essentially claiming that all vector bundles on $S^2$ are trivial and this is false, e.g. the tangent bundle is nontrivial (this is related to the Poincare-Hopf theorem) because its Euler class is nonvanishing.