短正合序列图

Question

有人能帮我吗？我想写出所有对象之间的态射

\[
  \setlength{\arraycolsep}{1pt}
  \begin{array}{*{9}c}
    0 &\Lrightarrow & X & \Lrightarrow & Y & \Lrightarrow & Z & \Lrightarrow & 0\\
    & & \Ldownarrow & & \Ldownarrow & & \Ldownarrow & & \\
    0 &\Lrightarrow & X^' & \Lrightarrow & Y^' & \Lrightarrow & Z^' &  \Lrightarrow & 0
  \end{array}
\]

Answer 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}

Answer 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}