% Simple 2d example illustrating the role of the Jacobian determinant in the change of variables formula. Inspired by Ari Seff in https://youtu.be/i7LjDvsLWCg?t=250.

\documentclass[tikz]{standalone}

\usepackage{mathtools}

\usetikzlibrary{calc,positioning,shapes.geometric}

\renewcommand\vec[1]{\boldsymbol{#1}}

\begin{document}
\begin{tikzpicture}[
    thick, node distance=15mm,
    set/.style={draw, diamond, text width=8mm, align=center},
    op/.style={draw, circle, text width=5mm, align=center, fill=orange!40},
  ]

  \node[set, fill=blue!20] (z1) {$\vec z_{1:d}$};
  \node[op, right=of z1] (eq) {\raisebox{-1ex}=};
  \node[set, right=of eq, fill=blue!20] (x1) {$\vec x_{1:d}$};
  \draw[->] (z1) edge (eq) (eq) edge (x1);

  \node[set, below=1 of z1, fill=green!30] (z2) {$\mathclap{\vec z_{d+1:D}}$};
  \node[op, right=of z2] (g) {$g$};
  \node[below=1em of g] (forward) {forward pass};
  \node[set, right=of g, fill=yellow!40] (x2) {$\mathclap{\vec x_{d+1:D}}$};
  \draw[->] (z2) edge (g) (g) edge (x2);

  \node[op] (m) at ($(z1)!0.5!(g)$) {$m$};
  \draw[->] (z1) edge (m) (m) edge (g);

  \begin{scope}[xshift=9cm]

    \node[set, fill=blue!20] (z1) {$\vec z_{1:d}$};
    \node[op, right=of z1] (eq) {\raisebox{-1ex}=};
    \node[set, right=of eq, fill=blue!20] (x1) {$\vec x_{1:d}$};
    \draw[<-] (z1) edge (eq) (eq) edge (x1);

    \node[set, below=1 of z1, fill=green!30] (z2) {$\mathclap{\vec z_{d+1:D}}$};
    \node[op, right=of z2] (g) {$\mathclap{g^{-1}}$};
    \node[below=1em of g] (inverse) {inverse pass};
    \node[set, right=of g, fill=yellow!40] (x2) {$\mathclap{\vec x_{d+1:D}}$};
    \draw[<-] (z2) edge (g) (g) edge (x2);

    \node[op] (m) at ($(x1)!0.5!(g)$) {$m$};
    \draw[->] (x1) edge (m) (m) edge (g);

  \end{scope}

\end{tikzpicture}
\end{document}
