1

I want to be like the output below. Please heip me! Thank you very much!

enter image description here

3

As I am guilty on up-voting Henri's comment, I repent by providing you:

% arara: pdflatex

\documentclass{article}
\usepackage{tikz-cd}

\begin{document}
\[
\begin{tikzcd}[column sep={2.3cm,between origins}]
& & & H_n(X_{n+1},X_n) = 0 & \\
H_n(X_{n-1}) = 0 \arrow{dr} & & H_n(X_{n+1}) \cong H_n(x) \arrow{ur} & & \\
& H_n(X_{n}) \arrow{dr}{j_n}\arrow{ur}{i_n} & & & \\
H_{n+1}(X_{n+1},X_n) \arrow{ur}{\partial_{n+1}} \arrow{rr}{d_{n+1}} & & H_n(X_{n},X_{n-1}) \arrow{dr}[swap]{\partial_n} \arrow{rr}{d_{n}} & & H_{n-1}(X_{n-1},X_{n-1}) \\
& & & H_{n-1}(X_{n-1}) \arrow{ur}[swap]{j_{n-1}} & \\
& & H_{n-1}(X_{n-2}) = 0 \arrow{ur} & & 
\end{tikzcd}
\]
\end{document}

enter image description here

2

A variation of LaRiFaRi's solution, for a more compact diagram:

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz-cd}

\begin{document}

\[
% the diagram has five columns and six rows
\begin{tikzcd}[column sep={5em,between origins}]
% row 1
&&& H_n(X_{n+1},X_n) \\
% row 2
H_n(X_{n-1})=0 \arrow[dr] && H_n(X_{n+1})\cong H_n(X) \arrow[ur] \\
% row 3
& H_n(X_n) \arrow[ur,"i_n"] \arrow[dr,"j_n"] \\
% row 4
H_{n+1}(X_{n+1},X_n) \arrow[ur,"\partial_{n+1}"] \arrow[rr,"d_{n+1}"] &&
  H_n(X_n,X_{n-1}) \arrow[rr,"d_n"] \arrow[dr,swap,"\partial_n"] &&[-3.5em]
  H_{n-1}(X_{n-1},X_{n-1}) \\
% row 5
&&& H_{n-1}(X_{n-1}) \arrow[ur,swap,"j_{n-1}"] \\
% row 6
&& H_{n-1}(X_{n-2}) \arrow[ur]
\end{tikzcd}
\]

\end{document}

enter image description here

  • @longhoang You are very welcome. Please accept one answer in order to close this post. Thanks. – LaRiFaRi Dec 19 '16 at 11:17

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