如何得到 $\Phi$ 下方图表底部的 $U$？

Question

\begin{center}
\begin{tikzpicture}

\node (E) at (0,0) {$E|_U:=\pi^{-1}(U)$};
\node[right=of E] (F) {$U\times \mathbb{R}^k$};
\node[below=of F] (A) {$U$};

\draw[->] (E)--(F) node [midway,above] {$\Phi$};
\draw[->] (E)--(A) node [midway,below] {$\pi$};
\draw[->] (F)--(A) node [midway,right] {$\pi_1$};


\end{tikzpicture}
\end{center}

另外如何在三角形中获得圆形箭头（图表通勤指示器）。

Answer 1

我认为我没有正确理解这个问题，但在托比昂的帮助下，事情开始明朗起来。可能是这样的？

\documentclass{article}
\usepackage{tikz}
\usepackage{amsfonts}
\usetikzlibrary{positioning}
\begin{document}
  \begin{center}
\begin{tikzpicture}

\node (E) at (0,0) {$E|_U:=\pi^{-1}(U)$};
\node[right=of E] (F) {$U\times \mathbb{R}^k$};

\draw[->] (E)--(F) node [midway,above] (phi) {$\Phi$};
\node[below= 1cm of phi] (A) {$U$};
\draw[->] (E)--(A.north west) node [midway,below] {$\pi$};
\draw[->] (F)--(A.north east) node [midway,right] {$\pi_1$};
\draw [<-] ([yshift=2mm]A.north) arc[start angle=-90,delta angle=350,radius=0.3cm];

\end{tikzpicture}
\end{center}
\end{document}

对于这样的图表，tikz-cd提供了友好的语法：

\documentclass{article}
\usepackage{tikz-cd,amsmath}
\usepackage{amsfonts}
\begin{document}
\[\begin{tikzcd}%[column sep=large]
   E|_U:=\pi^{-1}(U)\arrow[rr,"\Phi" ]\arrow[dr,swap,"\pi" ]& & U\times \mathbb{R}^k \arrow[dl,"\pi_1" ] \\
      &  U &
\end{tikzcd}
\]
\end{document}