我想把这个图表纳入我的工作中
但是，我不知道如何开始，因为我从未使用过 tikz，而是使用更普通的 tikzcd 来制作交换图。我应该指出三件事，我绝对不知道如何将它们引入任何类型的图表中：
- 在图中，出现了一个圆圈，里面有一个 1。它应该出现在它所在的位置，并且按照它原来的位置出现。
- 箭头
- 如果您注意到，与两个立方体相接的面的元素会有一些倾斜，因此它们与相应的线平行。模仿这种效果会很酷。
我知道公开提出问题的人尝试过的一些代码是习惯性的，但就我而言，我对 tikz 完全没有经验......
很抱歉造成您的不便，谢谢。
答案1
第一次尝试。代码不太好，但可以使用3d库。
\documentclass{standalone}
\newcommand{\Tcall}{\mathcal{T}}
\newcommand{\Scall}{\mathcal{S}}
\usepackage{tikz}
\usetikzlibrary{3d,positioning}
\begin{document}
\begin{tikzpicture}[z={(10:10mm)},x={(-45:5mm)}]
\def\step{2cm}
\begin{scope}[canvas is zy plane at x=-2]
\node(fGs){$\Gamma(\overline{\Scall})$};
\node[right=\step of fGs,transform shape](H1MS) {$H^1(M,\Scall)$};
\node[right=\step of H1MS,transform shape] (f0) {$0$};
\node[right=\step of f0,transform shape] (v1) {};
\node[below=\step of fGs](fG) {$\Gamma(\Scall)$};
\node[right=\step of fG,transform shape](H1MSs) {$\tilde{H}^1(M,\Scall)$};
\node[right=\step of H1MSs,transform shape] (s0) {$0$};
\node[right=\step of s0,transform shape] (v2) {};
\draw[->,transform shape] (fGs) -- (H1MS);
\draw[->,transform shape] (H1MS) -- (f0);
\draw (f0) -- (v1);
\draw[->,transform shape] (fGs) -- (fG);
\draw[->,transform shape] (fG) -- (H1MSs);
\draw[->,transform shape] (H1MSs) -- (s0);
\draw (s0) -- (v2);
\draw[->,transform shape] (H1MS) -- (H1MSs);
\draw[->,transform shape] (f0) -- (s0);
\end{scope}
\begin{scope}[canvas is zy plane at x=2]
\node(fIs){$\Gamma(\overline{\Tcall})$};
\node[right=\step of fIs,transform shape](H1MI) {$H^1(M,\Tcall)$};
\node[right=\step of H1MI,transform shape] (t0) {$0$};
\node[right=\step of t0,transform shape] (v3) {};
\node[below=\step of fIs](fI) {$\Gamma(\Tcall)$};
\node[right=\step of fI,transform shape](H1MIs) {$\tilde{H}^1(M,\Tcall)$};
\node[right=\step of H1MIs,transform shape] (fo0) {$0$};
\node[right=\step of fo0,transform shape] (v4) {};
\draw[->,transform shape] (fIs) -- (H1MI);
\draw[->,transform shape] (H1MI) -- (t0);
\draw (t0) -- (v3);
\draw[->,transform shape] (fIs) -- (fI);
\draw[->,transform shape] (fI) -- (H1MIs);
\draw[->,transform shape] (H1MIs) -- (fo0);
\draw (fo0) -- (v4);
\draw[->,transform shape] (H1MI) -- (H1MIs);
\draw[->,transform shape] (t0) -- (fo0);
\end{scope}
\draw[->] (fGs) -- (fIs);
\draw[->] (fG) -- (fI);
\draw[->] (H1MS) -- (H1MI);
\draw[->] (H1MSs) -- (H1MIs);
\draw[->] (f0) -- (t0);
\draw[->] (s0) -- (fo0);
\path (H1MS) -- (H1MIs) node[midway, circle, inner sep=0pt,draw] {1};
\end{tikzpicture}
\end{document}
有一些小进步
\documentclass{standalone}
\newcommand{\Tcall}{\mathcal{T}}
\newcommand{\Scall}{\mathcal{S}}
\usepackage{tikz}
\usetikzlibrary{3d,positioning,decorations.markings}
\begin{document}
\begin{tikzpicture}[z={(10:10mm)},x={(-45:5mm)},
decoration={markings,
mark=at position .65 with {\arrow[transform shape]{latex}}},
conn/.style={thick,postaction={decorate}}]
\def\step{2cm}
\begin{scope}[canvas is zy plane at x=-2]
\node(fGs){$\Gamma(\overline{\Scall})$};
\node[right=\step of fGs,transform shape](H1MS) {$H^1(M,\Scall)$};
\node[right=\step of H1MS,transform shape] (f0) {$0$};
\node[right=\step of f0,transform shape] (v1) {};
\node[below=\step of fGs](fG) {$\Gamma(\Scall)$};
\node[right=\step of fG,transform shape](H1MSs) {$\tilde{H}^1(M,\Scall)$};
\node[right=\step of H1MSs,transform shape] (s0) {$0$};
\node[right=\step of s0,transform shape] (v2) {};
\draw[conn] (fGs) -- (H1MS);
\draw[conn] (H1MS) -- (f0);
\draw[thick] (f0) -- (v1);
\draw[conn] (fGs) -- (fG);
\draw[conn] (fG) -- (H1MSs);
\draw[conn] (H1MSs) -- (s0);
\draw[thick] (s0) -- (v2);
\draw[conn] (H1MS) -- (H1MSs);
\draw[thick] (f0) -- (s0);
\end{scope}
\begin{scope}[canvas is zy plane at x=2]
\node(fIs){$\Gamma(\overline{\Tcall})$};
\node[right=\step of fIs,transform shape](H1MI) {$H^1(M,\Tcall)$};
\node[right=\step of H1MI,transform shape] (t0) {$0$};
\node[right=\step of t0,transform shape] (v3) {};
\node[below=\step of fIs](fI) {$\Gamma(\Tcall)$};
\node[right=\step of fI,transform shape](H1MIs) {$\tilde{H}^1(M,\Tcall)$};
\node[right=\step of H1MIs,transform shape] (fo0) {$0$};
\node[right=\step of fo0,transform shape] (v4) {};
\draw[conn] (fIs) -- (H1MI);
\draw[conn] (H1MI) -- (t0);
\draw[thick] (t0) -- (v3);
\draw[conn] (fIs) -- (fI);
\draw[conn] (fI) -- (H1MIs);
\draw[conn] (H1MIs) -- (fo0);
\draw[thick] (fo0) -- (v4);
\draw[conn] (H1MI) -- (H1MIs);
\draw[thick] (t0) -- (fo0);
\end{scope}
\draw[conn] (fGs) -- (fIs);
\draw[conn] (fG) -- (fI);
\draw[conn] (H1MS) -- (H1MI);
\draw[conn] (H1MSs) -- (H1MIs);
\draw[thick] (f0) -- (t0);
\draw[thick] (s0) -- (fo0);
\path (H1MS) -- (H1MIs) node[midway, circle, inner sep=0pt,draw] {1};
\end{tikzpicture}
\end{document}
答案2
pst-node和的可能性pst-arrow：
\documentclass{article}
\usepackage{fourier}
\usepackage{pst-node, pst-arrow}
\usepackage{auto-pst-pdf}
\begin{document}
\[ \begin{psmatrix}[rowsep=0.6cm, colsep=1.7cm]
& & & [name=0E] 0 & \pnode[0,2.7ex]{IE}\\
[name=GE]\Gamma(\bar{\mathcal{E}}) & & & & [name=0T] 0 & \pnode[0,2.7ex]{IT}\\
& [name=GT]\Gamma(\,\overline{\!\mathcal{T}\!}\,) \\
& & & [name=0Eb] 0 & \pnode[0,2.7ex]{IEb} \\
[name=GEb] \Gamma(\bar{\mathcal{E}}) & & & & [name=0Tb]0 & \pnode[0,2.7ex]{ITb} \\
& [name=GTb] \Gamma(\,\overline{\!\mathcal{T}\!}\,)
\end{psmatrix}
{\psset{linestyle=none, nrot=:U}%
\ncline{GE}{0E}\ncput*{\rnode{HE}{H^1(M, \mathcal{E})}}
\ncline{GT}{0T}\ncput*{\rnode{HT}{H^1(M, \mathcal{T})}}
\ncline{GEb}{0Eb}\ncput*{\rnode{HEb}{\tilde{H}^1(M, \mathcal{E})}}
\ncline{GTb}{0Tb}\ncput*{\rnode{HTb}{\tilde{H}^1(M, \mathcal{T})}}
}%
\ncline{0E}{0T}\ncline{0T}{0Tb}\ncline{0Tb}{0Eb}\ncline{0E}{0Eb}
{\psset{nodesepA=1pt} %
\ncline{0E}{IE}\ncline{0T}{IT} \ncline{0Eb}{IEb}\ncline{0Tb}{ITb}}%
{\psset{arrowinset=0.12, ArrowInside=->, nodesepB=3pt}%
\ncline{HE}{0E}
\ncline{HT}{0T}
\ncline{HEb}{0Eb}
\ncline{HTb}{0Tb}
\psset{ArrowInsidePos=0.66, }
\ncline{GE}{GEb}\ncline{GEb}{GTb}
\ncline{GE}{GT}\ncline{GT}{GTb}
\ncline{HE}{HT}\ncline{HT}{HTb}
\ncline{HE}{HEb}\ncline{HEb}{HTb}
}
\psset{arrows=->}
\ncline{GE}{HE}\ncline{GT}{HT}
\ncline{GEb}{HEb}\ncline{GTb}{HTb}
%
\]
\end{document}
