The universal complex flag bundle is BT-n

Write $V=V_n({\mathbb{C}}^{\mathrm{\infty }})$ and $Q=\mathrm{Fl}({\gamma }_n)$. A frame $(v_1,\dots ,v_n)$ determines the flag spanned by its first $i$ vectors. Two frames give the same flag precisely when their individual vectors differ by unit scalars, so this identifies $Q$ with $V\mathrm{/}{T}^n$ over the Grassmannian. This is a topological identification: in a Grassmannian graph chart, Gram–Schmidt identifies the frame projection with $U\times U(n)\to U$; the flag construction in [F8] identifies the corresponding flag chart with $U\times U(n)\mathrm{/}{T}^n$. These identifications agree on overlaps. The stable graph formulas are continuous on each finite stage, and their inverses remain continuous after multiplying by the compact group $U(n)$, so give the ordinary stable bundle charts. Equivalently, over a flag chart choose a nonzero local section of each orthogonal line and normalize it; the resulting unit vectors give local sections of $V\to Q$. Thus this is a principal $T^n$-bundle, and $L_i$ is its $i$-th coordinate line. The Grassmannian is a paracompact Hausdorff CW complex by [F9], its tautological charts are numerable by [F9], and [F8] supplies that $Q$ is paracompact Hausdorff CGWH of CW type. Applying [F9] to the flag chart cover makes $V\to Q$ numerable.

Let $P=({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}{)}^n$ with ordinary product topology and let $A=(S^{\mathrm{\infty }}{)}^n\to P$ be the product of the standard circle bundles. These are the circle models of [F2]. A product of their finitely many local charts is an ordinary principal $T^n$-chart; multiplying the finitely many partition functions gives a support-subordinate locally finite numeration. Their total product is contractible, by taking the product of their contractions. Moreover this product is a classifying bundle, without assuming that assertion from contractibility: a numerable principal $T^n$-bundle $D\to X$ gives the $n$ circle bundles $D\mathrm{/}{K}_i$, where $K_i$ is the kernel of the $i$-th coordinate homomorphism. Its charts and numeration descend to each quotient. The map $D\to {\prod }_X(D\mathrm{/}{K}_i)$ is a bundle isomorphism, as is seen in every principal chart, where it is the identity of $(S^1{)}^n$. Conversely a finite collection of numerable circle bundles has a numerable fiber product by multiplying partitions. These two constructions are inverse on isomorphism classes. By [F3] for $S^1$, such classes are therefore naturally ${\prod }_i[X,{\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}]=[X,P]$; the equality holds since maps and homotopies into finite products are coordinatewise. Thus $P$ is a model of $B{T}^n$.

The ordinary space $P$ is a CW complex. Here the countability qualification matters: each factor has countably many cells by [F9], so the finite product CW structure has the ordinary product topology. One can check the latter directly by exhausting each of two countable CW complexes by finite subcomplexes $X_j,Y_j$. If $W$ is open in the product cell topology and $(a,b)\in W$, start with a compact product neighborhood $K_1\times M_1\subset W$ in $X_1\times Y_1$. Inductively choose compact neighborhoods $K_{j+1}$ of $K_j$ and $M_{j+1}$ of $M_j$ in the next finite stages with $K_{j+1}\times M_{j+1}\subset W$: compactness first gives product neighborhoods at each point of $K_j$, then a finite subcover gives the union in the first coordinate and intersection in the second. The unions of their interiors are open by the weak topologies and their product lies in $W$. This proves equality of the product and cell topologies; iterate finitely. Closure finiteness and the cell characteristic maps follow from products of the finite-stage cells.

Send a frame to its ordered unit vectors, obtaining a continuous $T^n$-equivariant map $V\to A$. It descends to $f:Q\to P$, sending a flag to its ordered orthogonal lines viewed as lines in ${\mathbb{C}}^{\mathrm{\infty }}$. The principal fiber map is the identity of $T^n$ after choosing corresponding basepoints. Both bundle projections are Serre fibrations by [F9]. Their total spaces are contractible by [F1] and step 1.2. For every $k\ge 2$, the connecting maps identify each base's ${\pi }_k$ with ${\pi }_{k-1}(T^n)$ by [F4], and naturality [F5] identifies $f_{\ast }$ with the identity through these isomorphisms. The low-degree exact sequence gives ${\pi }_1=0$ since $T^n$ is path connected. Both bases are path connected as images of their contractible total spaces, so $f$ is a weak homotopy equivalence in every degree. To apply [F6] correctly to the CW-type space $Q$, choose a homotopy equivalence $h:C\to Q$ with $C$ CW. The composite $fh:C\to P$ is a weak equivalence of CW complexes, hence a homotopy equivalence. Since $h$ is also a homotopy equivalence, so is $f$. The bundle over $Q$ is the pullback of the classifying product bundle along $f$, hence is itself classifying: precomposition with a homotopy equivalence gives bijections $[X,Q]\cong [X,P]$ for every $X$. This licenses the quotient model $Q=B{T}^n$ over $BU(n)$.

Explicitly, the $i$-th coordinate of $f$ is the line $L_i$ itself, so $L_i$ is the pullback of the standard tautological line on the $i$-th factor. By [F10], $t_i=f^{\ast }{\mathrm{pr}}_i^{\ast }u$ with the tautological Euler generator $u$ of [F7]; no sign change to the dual-line convention is made. Iterating [F7] is valid because a projective-space factor has finite free integral homology in each degree. It gives $H^{\ast }(P;\mathbb{Z})=\mathbb{Z}[u_1,\dots ,u_n]$, and the homotopy equivalence $f$ gives the asserted ring on $Q$.

Permutation matrices normalize $T^n$. Right multiplication therefore descends from $V$ to homeomorphisms of $Q$ covering the identity on $BU(n)$; it need not be an equivariant automorphism of the original principal $U(n)$-bundle. On the ordered orthogonal lines it is the corresponding permutation, and $f$ intertwines this action with permutation of the coordinates of $P$. Thus it permutes the $t_i$. For any $a\in H^{\ast }(BU(n);\mathbb{Z})$ and any such homeomorphism $\sigma$, the equality $q\sigma =q$ gives ${\sigma }^{\ast }{q}^{\ast }a=q^{\ast }a$. The image is therefore contained in the symmetric invariants, as claimed.

For $n=1$, there are no flag-construction steps: $Q=BU(1)={\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}$, $f$ is the identity, the sole line is the universal line and the permutation group is trivial. The hypothesis $n\ge 1$ excludes rank zero; these universal bases are nonempty. All products are finite and $\mathbb{Z}$ is nonzero. AC is inherited from classification, partitions, Whitehead and Kunneth, not from any finite choice of coordinates.