# commutative Diagram in LaTex

I want to be like the output below. Please heip me! Thank you very much! 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} 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} • @longhoang You are very welcome. Please accept one answer in order to close this post. Thanks. Dec 19 '16 at 11:17