9

This question almost duplicates Bending snakes with xy-pic, but I need something slightly different from the solutions presented there. I spent some time trying to figure out how arrow bending works in XY-pic, but I don't seem to understand the syntax.

So I have the following diagram:

\[ \xymatrix@M=1em{
 & \ker d^\prime\ar@{>->}[d]\ar[r] & \ker d\ar@{>->}[d]\ar[r] & \ker d^{\prime\prime}\ar@{>->}[d]\ar[dddll] \\
 & M^\prime\ar[r]\ar[d] & M\ar@{->>}[r]\ar[d] & M^{\prime\prime}\ar[r]\ar[d] & 0 \\
0\ar[r] & N^\prime\ar@{>->}[r]\ar@{->>}[d] & N\ar[r]\ar@{->>}[d] & N^{\prime\prime}\ar@{->>}[d] \\
 & \coker d^\prime\ar[r] & \coker d\ar[r] & \coker d^{\prime\prime}} \]

the diagram

I want to bend the diagonal arrow so that it looks like the red arrow I painted above: goes down, crosses the arrow M'' -> 0, then crosses the three vertical arrows, and finally the arrow 0 -> N'. I want it to consist of straight segments, with slightly rounded corners. I am pretty sure the "bending" feature of XY-pic should do that, but I could not produce the desired result following some examples.

I found another question: How do you draw the "snake" arrow for the connecting homomorphism in the snake lemma?, with the accepted solution that looks almost as what I want, but I don't want to use TikZ. I hope someone here can show me how XY-pic bending is done in this case.

Thank you!

9
+150

A handy (and short) document is the XY-pic User's Guide by Kristoffer H. Rose, available as "User Guide" at https://www.ctan.org/pkg/xypic?lang=en. Curving the arrows is section 2.8, and the best you'd get is something like:

\documentclass{article}
\usepackage[arrow,matrix]{xy}
\newcommand{\coker}{\mathrm{coker}~}
\begin{document}
\[
 \xymatrix@M=1em{
 & \ker d'\ar@{>->}[d]\ar[r] & \ker d\ar@{>->}[d]\ar[r] & \ker d''\ar@{>->}[d]
 \ar`r[dr]`[dlll]`[dddlll]`[dddll][dddll] %\ar[dddll]
 \\
 & M'\ar[r]\ar[d] & M\ar@{->>}[r]\ar[d] & M''\ar[r]\ar[d] & 0 \\
 0\ar[r] & N'\ar@{>->}[r]\ar@{->>}[d] & N\ar[r]\ar@{->>}[d] & N''\ar@{->>}[d] \\
 & \coker d'\ar[r] & \coker d\ar[r] & \coker d''}
\]
\end{document}

(I've replace \primeskip with ') which becomes:

first attempt at diagram

Let's breakdown the command of interest:

\ar       % an arrow
 `r       % which starts by going to the right before curving to go toward
 [dr]     % the entry one down and one to the right
 `        % before bending to go toward
 [dlll]   % the entry one down and three to the left
 `        % before bending to go toward
 [dddlll] % the entry three down and three to the left
 `        % before bending to go toward
 [dddll]  % the entry three down and two to the left
 [dddll]  % where the arrow ends

To get something closer to what you're looking for, we can take a trick from section 2.7 and define some phantom targets. We'll put in extra columns and rows where the arrow would need to curve. To make the spacing come out right, we'll also put in extra columns and rows everywhere else. So instead of 5 columns and 4 rows, we'll have 9 columns and 7 rows (but every other column / row won't have an entry; a few of them are just targets for the arrow to curve at). That spaces things out too much, so we tighten up the sizing (a lot of guess and check to match the original sizing). Finally, instead of [ddddddlllll] as a target, we'll use the relative [6,-5] (see section 2.5). That leaves us with:

\documentclass{article}
\usepackage[arrow,matrix]{xy}
\newcommand{\coker}{\mathrm{coker}~}
\begin{document}
\[
 \xymatrix@C=.3em@R=.7ex@M=1em{
  && \ker d'\ar@{>->}[dd]\ar[rr] && \ker d\ar@{>->}[dd]\ar[rr] && \ker d''\ar@{>->}[dd]
  \ar`r[3,1]`[3,-5]`[6,-5]`[6,-4][6,-4]
  && \\\\
  && M'\ar[rr]\ar[dd] && M\ar@{->>}[rr]\ar[dd] && M''\ar[rr]\ar[dd] && 0 \\
  &&&&&&&& \\ % needs enough columns to have the target we want
  0\ar[rr] && N'\ar@{>->}[rr]\ar@{->>}[dd] && N\ar[rr]\ar@{->>}[dd] && N''\ar@{->>}[dd] \\\\
  && \coker d'\ar[rr] && \coker d\ar[rr] && \coker d''}
\]
\end{document}

and

final diagram

  • That's exactly what I was looking for, thank you very much for the solution and explanation. – user45435 Jul 1 '16 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy