# Commutative diagram: get an arrow to 'skip' an object

    $$\begin{tikzcd} A\ar[r,shift left,"f"] &\ar[l,shift left,"h"] B\ar[r,shift left,"g"] &\ar[l,shift left,"l"]\ar[l,bend left, shift left,"h"] C\end{tikzcd}$$


This is almost as I'd like it to be, except that the $h$ arrow is going from $C$ to $B$ rather than from $C$ to $A$. How can I modify the code so it does that?

EDIT: accidentally called two arrows 'h'. This was not intentional, but I think the text makes it clear which one I'm referring to.

Use ll in \ar[ll,bend left, shift left,"h"]. Two ls mean that the arrow should go two positions to the left.

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

\begin{document}
$\begin{tikzcd} A\ar[r,shift left,"f"] &\ar[l,shift left,"h"] B\ar[r,shift left,"g"] &\ar[l,shift left,"l"]\ar[ll,bend left, shift left,"h"] C \end{tikzcd}$
\end{document}


• And in LaTeX, don't use $$...$$, use $...$ as the saying goes....
– user11232
Oct 18, 2015 at 12:09
• That's for the answer. But what's the advantage of [ \backslash [...\backslash ]]? Oct 18, 2015 at 12:25
• @man_in_green_shirt: Please see: Why is $...$ preferable to $$...$$?
– user11232
Oct 18, 2015 at 12:31
• Had never heard of that. Thanks, will be sure to use it in the future! Oct 18, 2015 at 12:35
• @man_in_green_shirt $ are plain TeX primitives. $f$ will be $$f$$ in LaTeX and $$f$$ will be $f$. However, $f$ is more widely used and doesn't have much drawbaacks as its $\$ cousin.
– user11232
Oct 18, 2015 at 13:48