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使用 TikZ-cd 绘制立方体

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使用 TikZ-cd 绘制立方体

我想用 TikZ-cd 画一个立方体。条目大小不同，这会导致与我的相同的问题前一个问题，因为我希望箭头平行。

 \[\begin{tikzcd}[row sep=25,column sep=0,nodes={inner sep=2pt},
      cells={nodes={align=center,text width=\widthof{$X_2 \times_{S_2} Y_2$}}}]
    & X_1 \times_{S_1} Y_1  \ar{rr} \ar{dd} \ar{dl} & &   Y_1  \ar{dd} \ar{dl} \\
    X_1 \ar[crossing over]{rr} \ar{dd} & & S_1 \\
      & X_2 \times_{S_2} Y_2  \ar{rr} \ar{dl} & &  Y_2  \ar{dl} \\
    X_2 \ar{rr} && S_2 \ar[from=uu,crossing over]
 \end{tikzcd}\]

在此处输入图片描述

如何实现箭头真正在节点处结束和开始？当然，它们在形式上确实在节点处开始和结束，但我的意思是，例如上箭头实际上在 Y_1 附近结束。

答案1

如果您不是使用边框之间的特定宽度来分隔列，而是使用原点（节点中心）之间的宽度来分隔列，那么看起来会更好。

% 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}

在此处输入图片描述

如果你想让它真正逼真，我建议添加一些透视调整：

% 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}

在此处输入图片描述

实际上，你甚至应该将它稍微旋转到等距视图中。我猜，Ti 还有其他解决方案钾Z 周围。但为了完整起见：

% 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}

在此处输入图片描述

答案2

您可以玩start anchor和end anchor。

梅威瑟：

\documentclass{article}
\usepackage{tikz-cd}
\begin{document}
 \[\begin{tikzcd}[row sep=25,column sep=0,nodes={inner sep=2pt},
      cells={nodes={align=center,text width=\widthof{$X_2 \times_{S_2} Y_2$}}}]
    & X_1 \times_{S_1} Y_1  \ar[end anchor={[xshift=15pt]}]{rr} \ar{dd} \ar{dl} & &   Y_1  \ar{dd} \ar{dl} \\
    X_1 \ar[crossing over,start anchor={[xshift=-15pt]},end anchor={[xshift=15pt]}]{rr} \ar{dd} & & S_1 \\
      & X_2 \times_{S_2} Y_2  \ar[end anchor={[xshift=15pt]}]{rr} \ar{dl} & &  Y_2  \ar{dl} \\
    X_2 \ar[start anchor={[xshift=-15pt]},end anchor={[xshift=15pt]}]{rr} && S_2 \ar[from=uu,crossing over]
 \end{tikzcd}\]
\end{document}

在此处输入图片描述

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