# Stable homotopy groups of unitary groups vs homotopy group of union of unitary groups

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 $$S^k$$.

• Can you give $U$ the structure of a CW complex where each $U(n)$ is a subcomplex? If so, then you can use cellular approximation. Mar 31, 2020 at 13:55
• I know that there are more sophisticated techniques to obtain a CW structure, however I do not know any elementary way of seeing this: In comparison to the way of obtaining the CW decomposition of the complex projective plane the sphere appears to be on the "wrong" side of the fibration. Mar 31, 2020 at 14:26
• I think you can use Matt E's answer here math.stackexchange.com/questions/285351/… to construct a CW structure on each $U(n)$ so that the natural inclusions are inclusions of a subcomplex. Mar 31, 2020 at 14:47

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.
• In the details of showing it is a contradiction, you are probably going to use the fact that a cofibration $X_n \rightarrow X$ is a neighborhood deformation retraction pair Apr 1, 2020 at 16:28