How to draw a commutative diagram?

Question

Is it possible to draw a (simple) commutative diagram using MathJax?

Amscd doesn't seem to work here on math.SE.

M(N)WE:

$$ \begin{CD} K(X) @>{ch}>> H(X;\mathbb Q);\\ @VVV @VVV \\ K(Y) @>{ch}>> H(Y;\mathbb Q); \end{CD} $$

Edit: [A.K. May 7, 2013]

As pointed elsewhere by Davide, this can be remedied now that MathJax 2.2 [beta] is deployed, by adding \require{AMScd}:

$$\require{AMScd} \begin{CD} K(X) @>{ch}>> H(X;\mathbb Q);\\ @VVV @VVV \\ K(Y) @>{ch}>> H(Y;\mathbb Q); \end{CD}$$

Answer 1

MathJax 2.2 beta was released recently, and it includes support for AMScd. I hope that we'll see this support on m.SE soon enough (recall that 2.1 beta was also tested on this site). We have Mathjax 2.2 beta deployed, although without AMScd support for now. Hopefully that too would be added soon enough.

While AMScd doesn't support diagonal arrows, it will make rectangular diagrams easier to draw.

Edit: As Davide Cervone points out, one can manually load AMScd by adding \require{AMScd} and using \begin{CD}...\end{CD}. One may want to consult the AMScd manual.

Answer 2

Here is the commutative diagram from Arturo's fix, touched up to use extensible arrows, and with some spacing tightened up a bit.

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} 0 & \ra{f_1} & A & \ra{f_2} & B & \ra{f_3} & C & \ra{f_4} & D & \ra{f_5} & 0 \\ \da{g_1} & & \da{g_2} & & \da{g_3} & & \da{g_4} & & \da{g_5} & & \da{g_6} \\ 0 & \ra{h_1} & 0 & \ra{h_2} & E & \ra{h_3} & F & \ra{h_4} & 0 & \ra{h_5} & 0 \\ \end{array} $$

The code is valid in both mathjax and latex. In latex, one should include the amsmath package to get extensible arrows, and I would also recommend less negative spacing (or use a better commutative diagram environment).

$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
0 & \ra{f_1} & A & \ra{f_2} & B & \ra{f_3} & C & \ra{f_4} & D & \ra{f_5} & 0 \\
\da{g_1} & & \da{g_2} & & \da{g_3} & & \da{g_4} & & \da{g_5} & & \da{g_6} \\
0 & \ra{h_1} & 0 & \ra{h_2} & E & \ra{h_3} & F & \ra{h_4} & 0 & \ra{h_5} & 0 \\
\end{array}
$$

Answer 3

It is possible to do (somewhat primitive) commutative diagrams using \array:

$$\begin{array}{ccccccccc} 0 & \xrightarrow{i} & A & \xrightarrow{f} & B & \xrightarrow{q} & C & \xrightarrow{d} & 0\\ \downarrow & \searrow & \downarrow & \nearrow & \downarrow & \searrow & \downarrow & \nearrow & \downarrow\\ 0 & \xrightarrow{j} & D & \xrightarrow{g} & E & \xrightarrow{r} & F & \xrightarrow{e} & 0\end{array}$$

Code:

\begin{array}{ccccccccc}
0 & \xrightarrow{i} & A & \xrightarrow{f} & B & \xrightarrow{q} & C & \xrightarrow{d} & 0\\

\downarrow & \searrow & \downarrow & \nearrow & \downarrow & \searrow & \downarrow & \nearrow & \downarrow\\

0 & \xrightarrow{j} & D & \xrightarrow{g} & E & \xrightarrow{r} & F & \xrightarrow{e} & 0
end{array}

I'm not sure it's possible to label diagonal arrows using this approach though.

Answer 4

I would modify Jack's answer slightly:

\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
0 & \ra{f_1} & A & \ra{f_2} & B & \ra{f_3} & C & \ra{f_4} & D & \ra{f_5} & 0 \\
\da{g_1} & & \da{g_2} & & \da{g_3} & & \da{g_4} & & \da{g_5} & & \da{g_6} \\
0 & \ras{h_1} & 0 & \ras{h_2} & E & \ras{h_3} & F & \ras{h_4} & 0 & \ras{h_5} & 0 \\
\end{array}

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 0 & \ra{f_1} & A & \ra{f_2} & B & \ra{f_3} & C & \ra{f_4} & D & \ra{f_5} & 0 \\ \da{g_1} & & \da{g_2} & & \da{g_3} & & \da{g_4} & & \da{g_5} & & \da{g_6} \\ 0 & \ras{h_1} & 0 & \ras{h_2} & E & \ras{h_3} & F & \ras{h_4} & 0 & \ras{h_5} & 0 \\ \end{array} $$

This would also work in true $\rm\LaTeX$ except for one thing: the \rlap{\scriptstyle#1} would need to be \rlap{$\scriptstyle#1$}.