If $X_n$ is the bouquet of n circles, then what is $K_0(X_n)$ and $K_1(X_n)$? I am super confused on how to approach this problem, since I've only seen $K_0$, $K_1$ computed for algebras, but I don't understand how the topological space $X_n$ is a C* algebra.
I imagine the solution would involve computing $K_0(\mathbb{R})$, $K_1(\mathbb{R})$ since $X_n$ is the one point compactification of $n$ copies of $\mathbb{R}$. $K_0(\mathbb{R}) = [\mathbb{R}]$ and $K_1(\mathbb{R}) = \frac{GL(\mathbb{R})}{[GL(\mathbb{R}), GL(\mathbb{R})]}$, so would $K_0(X_n) = \Pi_{i=1}^{n}[\mathbb{R}]$, $K_1(X_n) = \Pi_{i=1}^{n}\frac{GL(\mathbb{R})}{[GL(\mathbb{R}), GL(\mathbb{R})]}$?
I assume you want to find the $K$-theory of $X_n$, not the $K$-homology of it (the $K$-theory groups of a space are typically denoted $K^i(X)$, not $K_i(X)$). For a compact Hausdorff space $X$, $K^i(X)$ is naturally isomorphic to $K_i(C(X))$, the (operator) $K$-theory of the $C^*$-algebra of complex-valued functions on $X$.
Note that we have a split exact sequence of algebras: $$0\to C_0\left(\bigsqcup^n_{i=1}\mathbb R\right)\to C(X_n)\leftrightarrows\mathbb C\to 0.$$ Thus, we get split exact sequences in both $K$-groups: $$0\to K_i\left(C_0\left(\bigsqcup^n_{i=1}\mathbb R\right)\right)\to K_i(C(X_n))\leftrightarrows K_i(\mathbb C)\to 0,$$ and so $$K_i(C(X_n))\cong K_i\left(C_0\left(\bigsqcup^n_{i=1}\mathbb R\right)\right)\oplus K_i(\mathbb C)\cong K_i(\mathbb C)\oplus\left(\bigoplus_{i=1}^nK_i(C_0(\mathbb R))\right)$$ Now we have $K_0(\mathbb C)=\mathbb Z$, $K_1(\mathbb C)=0$, $K_0(C_0(\mathbb R))=0$, and $K_1(C_0(\mathbb R))=\mathbb Z$, and therefore \begin{align*} K^0(X_n)&\cong K_0(C(X_n))\cong\mathbb Z\\ K^1(X_n)&\cong K_1(C(X_n))\cong \mathbb Z^n \end{align*}