\,
        2}}\parT{\uv}~,  && \quad \uv \in \CySpTT^{m, \, 2}~,\\ 
&\Pibr^3 u && = \parT{\eta_{m, \, 3}^{-1} \comp
    \Pi^2\comp \eta_{m, \, 3}}\parT{u} = \Pi^2\parT{u}~,
&&\quad u \in \CySpTT^{m, \, 3}~. 
\end{alignedat}$$ The following lemma shows that they are actually projectors.

The interpolators are projectors, that is $$\begin{aligned}
{3}
    &\Pibr^0 u_h && = u_h~, && \qquad u_h \in
    \CySpT_h^0~, \label{eq:proj_p0}\\ 
    &\Pibr^1 \uv_h && = \uv_h~, && \qquad\uv_h \in
    \CySpT_h^1~,\label{eq:proj_p1}\\ 
    &\Pibr^2 \uv_h && = \uv_h~, && \qquad\uv_h \in
    \CySpT_h^2~,\label{eq:proj_p2}\\ 
    &\Pibr^3 u_h && = u_h~, && \qquad u_h \in
    \CySpT_h^3~.\label{eq:proj_p3} 
    \end{aligned}$$

For the result is immediate since the projector coincides with the one defined on the Cartesian space. Also for – the result is trivial and due to the definition of the spaces and to the result in Cartesian coordinates. Taking, for example , $u_h \in \CySpT_h^0$, we have that $$\Pibr^0 u_h = \Pi^0\parT{\eta_{m, \, 0}\parT{u_h}} =
    \Pi^0\parT{\ut_h} = \ut_h~, \qquad\ut_h \in \CaSpB_h^0~.$$

We show that the continuous and discrete spaces, together with the projectors, form a commutative diagram. We start proving the commutativity of the top part of the diagram .

\[lemma:comm\_top\] The following diagram commutes: $$\begin{split}
    \tikzsetnextfilename{de_rham/de_rham_3d_cyl_top_tilde}
    \begin{tikzpicture}[commutative diagrams/every diagram,
    arr/.style={commutative diagrams/.cd,
        every arrow,
        every label}]
    \def\minW{25mm}
    \def\minH{18mm}
    \def\hShift{10mm}
    \def\vShiftA{20mm}
    \def\vShiftB{4mm}
    \def\auxA{1mm}
    