The QR and polar maps are first-order Stiefel and Grassmann retractions

On the full-rank set, the Gram–Schmidt recurrences divide only by positive residual norms, so [F3] expresses the QR factors by smooth additions, products, divisions by nonzero functions, and positive square roots. The polar formula is smooth there by [F4], [F5] applied to its positive-definite Gram matrix, and smooth matrix multiplication. Both maps land in Stiefel by [F1]. At $\eta =0$, the factorizations are $X=X{I}_r$, so each map returns $X$.

For the polar map, $Y(t{)}^{\mathsf{T}}Y(t)=I_r+t^2{\eta }^{\mathsf{T}}\eta$ by [F2]. Its inverse square root equals $I_r+O(t^2)$ by smoothness in [F4] and [F5] and its value $I_r$ at $I_r$. Consequently $R_X^{\mathrm{p}\mathrm{o}\mathrm{l}}(t\eta )=X+t\eta +O(t^2)$, so its derivative at zero is $\eta$.

Set $Y(t)=X+t\eta$ and write its positive-diagonal QR factorization as $Y(t)=Q(t)R(t)$. By step 1.1, $Q,R$ are differentiable near zero; $Q(0)=X$, $R(0)=I_r$. Put $W=Q^{\mathrm{\prime }}(0)$ and $U=R^{\mathrm{\prime }}(0)$, so $\eta =W+XU$ and $U$ is upper triangular. Differentiating $Q(t{)}^{\mathsf{T}}Q(t)=I_r$ shows $X^{\mathsf{T}}W$ is skew-symmetric. By [F2], $X^{\mathsf{T}}\eta$ is skew-symmetric; hence $U=X^{\mathsf{T}}\eta -X^{\mathsf{T}}W$ is skew-symmetric as well. A real matrix both upper triangular and skew-symmetric is zero, so $U=0$ and $W=\eta$.

Under $(X,\eta )\mapsto (X{Q}_0,\eta Q_0)$, the inputs $X+\eta$ and $(X+\eta )Q_0$ have the same column space. Their QR outputs may differ as frames, but induce the same Grassmann point; the polar output transforms by $Q_0$. Differentiating the induced maps therefore gives the same horizontal class of $\eta$, which is the identity on the quotient tangent space.