$GL(n,R)\cong O(n,R)\times B(n,R) $ as smooth manifolds, but are they isomorphic as Lie groups?

Question

In the title, $B(n,R)$ is the upper tiangular matrix with positive entries on the diagonal. According to Gram-Schmidt normalization, we know that $\forall G\in GL(n,R),\exists U\in O(n,R),T\in B(n,R)$，s.t. $G=UT$,the corresponding $U,T$ are unique, which induces diffeomorphism between $GL(n,R)$and $O(n,R)\times B(n,R)$ as smooth manifolds. My question is , does there exists a Lie group homeomriphism $f: GL(n,R)\rightarrow O(n,R)\times B(n,R)$?

The polar decomposition does not give a Lie group homeomorphism naturally because it does not preserves multiplicaton of matrix, i.e. $G_i=U_iT_i,i=1,2$，but $G_1G_2=U_1U_2T_1T_2$ does not hold for all elements in $GL(n,R)$.

Answer 1

No, the derived group $[Gl(n,R),Gl(n,R)]=Sl(n,R)$ of $Gl(n,R)$ is semisimple not the derived group of $O(n,R)\times B(n,R)$.