Projected vector transports return tangent representatives

Put $S=\mathrm{sym}(Y^{\mathsf{T}}Z)$ and $W=Z-YS$. Then $Y^{\mathsf{T}}W=Y^{\mathsf{T}}Z-S$ is skew-symmetric, so $Y^{\mathsf{T}}W+W^{\mathsf{T}}Y=0$. By [F2], $W$ is a Stiefel tangent. The projection formulas are linear in $Z$ and smooth in $Y$; [F3] makes their composition with the retraction smooth in $\eta$.

The Grassmann output $V=(I-Y{Y}^{\mathsf{T}})Z$ satisfies $Y^{\mathsf{T}}V=0$, so it is horizontal by [F2], and it depends linearly on $Z$. Under $(X,\eta ,Z)\mapsto (XQ,\eta Q,ZQ)$, [F4] makes the polar endpoint $Y$ change to $YQ$. Then $I-(YQ)(YQ{)}^{\mathsf{T}}=I-Y{Y}^{\mathsf{T}}$ and the output changes to $VQ$, exactly the changed horizontal representative. This proves independence of the represented tangent vector.

At $\eta =0$, [F3] gives $Y=X$. If $Z$ is Stiefel tangent at $X$, $X^{\mathsf{T}}Z$ is skew-symmetric, so $\mathrm{sym}(X^{\mathsf{T}}Z)=0$ and the Stiefel transport returns $Z$. If $Z$ is horizontal, $X^{\mathsf{T}}Z=0$ and $(I-X{X}^{\mathsf{T}})Z=Z$. Thus both transports are identity at the zero step, completing the vector-transport conditions.