[Tex/LaTex] Adjoint functors with tikz-cd

tikz-cd

I'd like to construct diagrams such as

enter image description here ,

preferably with tikz-cd, but I've no idea where to start. Please help 🙂

EDIT: This is what I have so far.

\begin{tikzcd}
\mathbf{C} \arrow[bend left=35]{r}[name=F]{F} & \mathbf{D}\arrow[bend left=35]{l}[name=U]{U}
\arrow[to path=(F) -- (U)]{d}{\vdash}
\end{tikzcd}

I've tried a few variations but I can't figure out how to get \vdash (or \bot , whichever).

Best Answer

A minor point on Qrrbrbirlbel's answer: the \dashv should be rotated. This can be done as follows:

\documentclass{standalone}
\usepackage{tikz-cd}
\usepackage{amsfonts}

\begin{document}
\begin{tikzcd}
\mathbb{C}
\arrow[r, "F"{name=F}, bend left=25] &
\mathbb{B}
\arrow[l, "U"{name=G}, bend left=25]
%--- Adjunction Symbol
\arrow[phantom, from=F, to=G, "\dashv" rotate=-90, no line]
\end{tikzcd}
\end{document}

giving

Related Solutions

[Tex/LaTex] tikz-cd error: Single ampersand used with wrong catcode

You missed one ampersand on the 2nd line.

Also, as observed by @egreg and the OP, the last & in the 1st and 3rd lines are not necessary, and you should start only the 1st line with {}.

\documentclass{amsart} 

\usepackage{tikz} 
\usepackage{tikz-cd}

\begin{document}
\[
\begin{tikzcd} 
{}& A \arrow{dl}\arrow{dr} & \\
B \arrow{dr} &&  C\arrow{dl} \\
  & D  
\end{tikzcd}
\]
\end{document}

enter image description here

[Tex/LaTex] Drawing a cube with TikZ-cd

If you separate your columns not with a certain width between borders, but with a width between origins (centre of nodes), it should look better.

% arara: lualatex

\documentclass{article}
\usepackage{tikz-cd}
\usepackage{lua-visual-debug} % just for proove of symmetry. Without that, you may compile with pdfLaTeX

\begin{document}
\[\begin{tikzcd}[row sep={40,between origins}, column sep={40,between origins}]
      & X_1 \times_{S_1} Y_1 \ar{rr}\ar{dd}\ar{dl} & & Y_1\vphantom{\times_{S_1}} \ar{dd}\ar{dl} \\
    X_1 \ar[crossing over]{rr} \ar{dd} & & S_1 \\
      & X_2 \times_{S_2} Y_2  \ar{rr} \ar{dl} & &  Y_2\vphantom{\times_{S_1}} \ar{dl} \\
    X_2 \ar{rr} && S_2 \ar[from=uu,crossing over]
\end{tikzcd}\]
\end{document}

enter image description here

If you want to have it really realistic, I would recommend to add some perspective tweaking:

% arara: pdflatex

\documentclass{article}
\usepackage{tikz-cd}
\newlength{\perspective}

\begin{document}
\setlength{\perspective}{2pt}
\[\begin{tikzcd}[row sep={40,between origins}, column sep={40,between origins}]
      &[-\perspective] X_1 \times_{S_1} Y_1 \ar{rr}\ar{dd}\ar{dl} &[\perspective] &[-\perspective] Y_1\vphantom{\times_{S_1}} \ar{dd}\ar{dl} \\[-\perspective]
    X_1 \ar[crossing over]{rr} \ar{dd} & & S_1 \\[\perspective]
      & X_2 \times_{S_2} Y_2  \ar{rr} \ar{dl} & &  Y_2\vphantom{\times_{S_1}} \ar{dl} \\[-\perspective]
    X_2 \ar{rr} && S_2 \ar[from=uu,crossing over]
\end{tikzcd}\]
\setlength{\perspective}{5pt}
\[\begin{tikzcd}[row sep={40,between origins}, column sep={40,between origins}]
      &[-\perspective] X_1 \times_{S_1} Y_1 \ar{rr}\ar{dd}\ar{dl} &[\perspective] &[-\perspective] Y_1\vphantom{\times_{S_1}} \ar{dd}\ar{dl} \\[-\perspective]
    X_1 \ar[crossing over]{rr} \ar{dd} & & S_1 \\[\perspective]
      & X_2 \times_{S_2} Y_2  \ar{rr} \ar{dl} & &  Y_2\vphantom{\times_{S_1}} \ar{dl} \\[-\perspective]
    X_2 \ar{rr} && S_2 \ar[from=uu,crossing over]
\end{tikzcd}\]
\setlength{\perspective}{8pt}
\[\begin{tikzcd}[row sep={40,between origins}, column sep={40,between origins}]
      &[-\perspective] X_1 \times_{S_1} Y_1 \ar{rr}\ar{dd}\ar{dl} &[\perspective] &[-\perspective] Y_1\vphantom{\times_{S_1}} \ar{dd}\ar{dl} \\[-\perspective]
    X_1 \ar[crossing over]{rr} \ar{dd} & & S_1 \\[\perspective]
      & X_2 \times_{S_2} Y_2  \ar{rr} \ar{dl} & &  Y_2\vphantom{\times_{S_1}} \ar{dl} \\[-\perspective]
    X_2 \ar{rr} && S_2 \ar[from=uu,crossing over]
\end{tikzcd}\]
\setlength{\perspective}{11pt}
\[\begin{tikzcd}[row sep={40,between origins}, column sep={40,between origins}]
      &[-\perspective] X_1 \times_{S_1} Y_1 \ar{rr}\ar{dd}\ar{dl} &[\perspective] &[-\perspective] Y_1\vphantom{\times_{S_1}} \ar{dd}\ar{dl} \\[-\perspective]
    X_1 \ar[crossing over]{rr} \ar{dd} & & S_1 \\[\perspective]
      & X_2 \times_{S_2} Y_2  \ar{rr} \ar{dl} & &  Y_2\vphantom{\times_{S_1}} \ar{dl} \\[-\perspective]
    X_2 \ar{rr} && S_2 \ar[from=uu,crossing over]
\end{tikzcd}\]
\end{document}

enter image description here

Actually, you should even rotate it a bit into isometric view. I guess, there are other solutions for TikZ around. But just to be complete:

% arara: pdflatex

\documentclass[twocolumn]{article}
\usepackage{tikz-cd}
\newlength{\perspective}

\begin{document}
    \setlength{\perspective}{12pt}
    \def\isofactor{0.5}
    \[\begin{tikzcd}[row sep={38,between origins}, column sep={38,between origins}]
    &[-\perspective] X_1 \times_{S_1} Y_1 \ar{rrd}\ar{dddd}\ar{ddl} &[\perspective] &[-\perspective] \\[-38+\isofactor\perspective]
    & & & Y_1\vphantom{\times_{S_1}} \ar{dddd}\ar{ddl} & \\[-\perspective-\isofactor\perspective]
    X_1 \ar[crossing over]{rrd} \ar{dddd} & & & \\[-38+\isofactor\perspective]
    & &  S_1 \\[\perspective-\isofactor\perspective]
    & X_2 \times_{S_2} Y_2  \ar{rrd} \ar{ddl} & &  \\[-38+\isofactor\perspective]
    & & & Y_2\vphantom{\times_{S_1}} \ar{ddl} \\[-\perspective-\isofactor\perspective]
    X_2 \ar{rrd} & & \\[-38+\isofactor\perspective]
    & & S_2 \ar[from=uuuu,crossing over] &
    \end{tikzcd}\]
    \setlength{\perspective}{10pt}
    \def\isofactor{1}
    \[\begin{tikzcd}[row sep={38,between origins}, column sep={38,between origins}]
    &[-\perspective] X_1 \times_{S_1} Y_1 \ar{rrd}\ar{dddd}\ar{ddl} &[\perspective] &[-\perspective] \\[-38+\isofactor\perspective]
    & & & Y_1\vphantom{\times_{S_1}} \ar{dddd}\ar{ddl} & \\[-\perspective-\isofactor\perspective]
    X_1 \ar[crossing over]{rrd} \ar{dddd} & & & \\[-38+\isofactor\perspective]
    & &  S_1 \\[\perspective-\isofactor\perspective]
    & X_2 \times_{S_2} Y_2  \ar{rrd} \ar{ddl} & &  \\[-38+\isofactor\perspective]
    & & & Y_2\vphantom{\times_{S_1}} \ar{ddl} \\[-\perspective-\isofactor\perspective]
    X_2 \ar{rrd} & & \\[-38+\isofactor\perspective]
    & & S_2 \ar[from=uuuu,crossing over] &
    \end{tikzcd}\]
    \vfill\break        
    \setlength{\perspective}{8pt}
    \def\isofactor{1.5}
    \[\begin{tikzcd}[row sep={38,between origins}, column sep={38,between origins}]
    &[-\perspective] X_1 \times_{S_1} Y_1 \ar{rrd}\ar{dddd}\ar{ddl} &[\perspective] &[-\perspective] \\[-38+\isofactor\perspective]
    & & & Y_1\vphantom{\times_{S_1}} \ar{dddd}\ar{ddl} & \\[-\perspective-\isofactor\perspective]
    X_1 \ar[crossing over]{rrd} \ar{dddd} & & & \\[-38+\isofactor\perspective]
    & &  S_1 \\[\perspective-\isofactor\perspective]
    & X_2 \times_{S_2} Y_2  \ar{rrd} \ar{ddl} & &  \\[-38+\isofactor\perspective]
    & & & Y_2\vphantom{\times_{S_1}} \ar{ddl} \\[-\perspective-\isofactor\perspective]
    X_2 \ar{rrd} & & \\[-38+\isofactor\perspective]
    & & S_2 \ar[from=uuuu,crossing over] &
    \end{tikzcd}\]
    \setlength{\perspective}{6pt}
    \def\isofactor{2}
    \[\begin{tikzcd}[row sep={38,between origins}, column sep={38,between origins}]
    &[-\perspective] X_1 \times_{S_1} Y_1 \ar{rrd}\ar{dddd}\ar{ddl} &[\perspective] &[-\perspective] \\[-38+\isofactor\perspective]
    & & & Y_1\vphantom{\times_{S_1}} \ar{dddd}\ar{ddl} & \\[-\perspective-\isofactor\perspective]
    X_1 \ar[crossing over]{rrd} \ar{dddd} & & & \\[-38+\isofactor\perspective]
    & &  S_1 \\[\perspective-\isofactor\perspective]
    & X_2 \times_{S_2} Y_2  \ar{rrd} \ar{ddl} & &  \\[-38+\isofactor\perspective]
    & & & Y_2\vphantom{\times_{S_1}} \ar{ddl} \\[-\perspective-\isofactor\perspective]
    X_2 \ar{rrd} & & \\[-40+\isofactor\perspective]
    & & S_2 \ar[from=uuuu,crossing over] &
    \end{tikzcd}\]
\end{document}

enter image description here

Best Answer

Related Solutions

[Tex/LaTex] tikz-cd error: Single ampersand used with wrong catcode

[Tex/LaTex] Drawing a cube with TikZ-cd

Related Question