有人能帮我吗?我想写出所有对象之间的态射
\[
\setlength{\arraycolsep}{1pt}
\begin{array}{*{9}c}
0 &\Lrightarrow & X & \Lrightarrow & Y & \Lrightarrow & Z & \Lrightarrow & 0\\
& & \Ldownarrow & & \Ldownarrow & & \Ldownarrow & & \\
0 &\Lrightarrow & X^' & \Lrightarrow & Y^' & \Lrightarrow & Z^' & \Lrightarrow & 0
\end{array}
\]
答案1
以下是xy
和的解tikz-cd
。想法是一样的,把它们当作矩阵来思考。
\documentclass{report}
\usepackage[all]{xy}
\usepackage{tikz-cd}
\begin{document}
\[
\xymatrix{
0 \ar[r] & A \ar[d]_-{\alpha} \ar[r]^-{f} & B \ar[d]_-{\beta} \ar[r]^-{g} & C \ar[d]^-{\gamma} \ar[r] & 0 \\
0 \ar[r] & A' \ar[r]_-{f'} & B' \ar[r]_-{g'} & C' \ar[r] & 0
}
\]
\[
\begin{tikzcd}
0 \arrow[r] & A \arrow[d, "\alpha"] \arrow[r, "f"] & B \arrow[d, "\beta"] \arrow[r, "g"] & C \arrow[d, "\gamma"] \arrow[r] & 0 \\
0 \arrow[r] & A' \arrow[r, "f'"] & B' \arrow[r, "g'"] & C' \ar[r] & 0
\end{tikzcd}
\]
\end{document}
答案2
解决方案如下psmatrix
:
\documentclass{article}
\usepackage{pst-node}
\usepackage{auto-pst-pdf}
\begin{document}
\[
\everypsbox{\scriptstyle}
\begin{psmatrix}[rowsep=1cm, colsep=1.2cm]
0 & A & B & C & 0 \\
0 & D & B' & C' & 0
%% Arrows
\psset{linewidth=0.5pt, arrows=->, arrowinset=0.12, nodesep=3pt, shortput=nab, labelsep=1.5pt}
\ncline{1,1}{1,2}\ncline{1,4}{1,5}\ncline{2,1}{2,2}\ncline{2,4}{2,5}
\ncline{1,2}{1,3}^{f}\ncline{1,2}{2,2}_{\alpha}
\ncline{1,3}{1,4}^{g}\ncline{1,3}{2,3}_{\beta}\ncline{1,4}{2,4}_{\gamma}
\ncline{2,2}{2,3}_{f'}
\ncline{2,3}{2,4}_{g'}
\end{psmatrix} \]%
\end{document}