There is a canonical inclusion $i$ of $U(n-1)$ into $U(n)$. Using the fibration $U(n-1)\rightarrow U(n)\rightarrow S^{2n-1}$ and the associated long exact sequence of homotopy one gets that $i_*$ induces an isomorphism $\pi_k(U(n-1))\cong\pi_k(U(n))$ for sufficiently large n. This allows one to form "stable" homotopy groups $\pi_k^s(U)$ of the unitary groups. (Note: "Stable" has a different meaning in comparison to "stable" homotopy groups of the spheres.)
Now the article on unitary groups on ncatlab claims that we can form the limit $U$ of the unitary groups to obtain the isomorphism $\pi_k^s(U)\cong \pi_k(U)$. My question is:
Is this construction valid? And, if yes, how do I see this?
Assume we are given a $f:S^k\rightarrow U$ representing $[f]\in \pi_k(U)$. I can't find a reason why $f$ should be homotopy-equivalent to a map $g$ with image contained in a $U(n)\subset U$. If the unitary groups $U(n)$ were open, this would follow from compactness of $S^k$.
Homotopy groups commute with "nice" filtered colimits. For example, if you have a direct limit of cofibrations, then the homotopy groups commute with the direct limit. This follows from the fact that the spheres are compact.
For your specific question, the image of $S^k$ is compact and so under some coherent CW decomposition (which these $U(n)$ have) it lies in finitely many cells. Necessarily, there must be some $U(n)$ that contains the entire image.