Why is $K_{0}(C(\mathbb{D}))\rightarrow K_{0}(C(\mathbb{T}))$ injective? There is the restriction map $\pi:C(\mathbb{D})\rightarrow C(\mathbb{T})$ where $\mathbb{D}$ is the closed unit disk and $\mathbb{T}$ is the unit circle. Why is $\pi_*:K_0(C(\mathbb{D}))\rightarrow K_0(C(\mathbb{T}))$ injective? Does it have to do with contractibility of $\mathbb{D}$?
 A: I'm more used to thinking of spaces rather than algebras, so I'll give an answer using vector bundles. You can probably translate it into the matrix setting yourself. From the long exact sequence for $K$-theory we have the following part, which is exact at $K^{0}(\mathbb{D})$:
$$ K^{0}(\mathbb{D},\mathbb{T}) \stackrel{j^{*}}{\longrightarrow} K^{{0}}(\mathbb{D}) \stackrel{i^{*}}{\longrightarrow} K^{0}(\mathbb{T}).$$ 
Now $K^{0}(\mathbb{D},\mathbb{T}) := \widetilde{K}^{0}(\mathbb{D} / \mathbb{T})
= \widetilde{K}^{0}(S^{2})$,
which is the kernel of the rank homomorphism 
$K^{0}(S^{2}) \rightarrow \mathbb{Z}$. In particular, a generic element of $\widetilde{K}^{0}(S^{2})$ is of the form 
$[\eta] - [\eta']$, where $\eta$ and $\eta'$ are any vector bundles on $S^{2}$ that have the same rank.
On the other hand, $\mathbb{D}$ is contractible, so the only vector bundles on it are the trivial bundles $\mathbb{D} \times \mathbb{C}^{n}$. The homomorphism $j^{*}: \widetilde{K}^{0}(S^{2}) \rightarrow K^{0}(\mathbb{D})$ sends vector bundles of rank $n$ to vector bundles of rank $n$ (as rank is preserved by pullback), so the image of both $\eta$ and $\eta'$ under $j^{*}$ is the trivial bundle $\mathbb{D} \times \mathbb{C}^{n}$. Hence $j^{*}$ is identically $0$ on $K^{0}(\mathbb{D}, \mathbb{T})$, and exactness of the sequence tells you that $i^{*} : K^{0}(\mathbb{D}) \rightarrow K^{0}(\mathbb{T})$ must be injective.
