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