Putting a hole at the right position (xypics)

Question

I need some help to put a hole at the right position please. Here is the diagram:

\xymatrix { \{\iota\} \sqcup A^0 \ar@{->}[rddd]\ar@{->}[dd]_-{\bigsqcup\limits_{\alpha\in A^0} g_\alpha} \ar@{->}[rr]^-{} && \{\iota\} \sqcup A \ar@{->}[rddd]^/20pt/{\rho^{\{\iota\}}(f)} \ar@{->}[rr]^-{\phi} \ar@{->}[dd]_-{} && X \ar@{->}[rdrdd]^-{g}  \\  && &&  \\ \bigsqcup\limits_{\alpha\in A^0} P_\alpha \ar@{->}[rddd]_-{\subset} \ar@{->}[rr]^-{}|!{[r]}\hole && \widehat{A}\ar@{-->}[rddd]_/10pt/{\exists !} \ar@{-->}[rruu]^-{\exists !} && \\ &\{\iota\} \sqcup B^0 \ar@{->}[dd]^-{\bigsqcup\limits_{\beta\in B^0} g_\beta} \ar@{->}[rr]^-{} && \{\iota\} \sqcup B \ar@{->}[rrr]^-{\psi} \ar@{->}[dd]_-{} &&& Y \\  &&& && \\ &\bigsqcup\limits_{\beta\in B^0} P_\beta \ar@{->}[rr]^-{} && \widehat{B} \ar@{-->}[rrruu]^-{\exists !} && }

I would like to push the hole on the left so that it is under the arrow from ${\iota} \sqcup A^0$ to ${\iota} \sqcup B^0$. I don't know how to do that. Thanks for your help.

score 4 · Accepted Answer · edited Apr 13 '17 at 12:35

In xy-matrix, you can specify how far you want the hole (or label) along the arrow using parentheses, so \ar[r]|(0.5)\hole places a hole halfway along the arrow, \ar[r]|(0.9)\hole places it nine-tenths of the way along the arrow, and so on. A little trial and error determined that in this case, it should be \ar@{->}[rr]^-{}|(0.315)\hole.

Here's the full diagram.

\xymatrix {
    \{\iota\} \sqcup A^0 \ar@{->}[rddd]
                         \ar@{->}[dd]_-{\bigsqcup\limits_{\alpha\in A^0} g_\alpha}
                         \ar@{->}[rr]^-{}
    && \{\iota\} \sqcup A \ar@{->}[rddd]^/20pt/{\rho^{\{\iota\}}(f)} 
                          \ar@{->}[rr]^-{\phi}
                          \ar@{->}[dd]_-{}
    && X \ar@{->}[rdrdd]^-{g} \\
    && && \\
    \bigsqcup\limits_{\alpha\in A^0} P_\alpha \ar@{->}[rddd]_-{\subset}
                                              \ar@{->}[rr]^-{}|(0.315)\hole % here
    && \widehat{A} \ar@{-->}[rddd]_/10pt/{\exists !}
                   \ar@{-->}[rruu]^-{\exists !} && \\
    &\{\iota\} \sqcup B^0 \ar@{->}[dd]^-{\bigsqcup\limits_{\beta\in B^0} g_\beta}
                          \ar@{->}[rr]^-{}
    && \{\iota\} \sqcup B \ar@{->}[rrr]^-{\psi}
                          \ar@{->}[dd]_-{} &&& \\
    &&& && \\
    &\bigsqcup\limits_{\beta\in B^0} P_\beta \ar@{->}[rr]^-{}
    && \widehat{B} \ar@{-->}[rrruu]^-{\exists !} &&
}

Related TeX.SX posts: 1 and 2.

(Finally, and this is a minor point, but \ar without @{} defaults to \ar@{->}, which could save you some typing. Similarly, you don't need to decorate with ^{} or _{}, since they don't do anything if empty.)

Thank you. Yes I knew for ^{} and _{} : they are in the source because I replaced the macros \fR{} and \fD{} I am using by what you mention in the source. — Philippe Gaucher, Oct 12 '15 at 16:12

score 3 · Answer 2 · answered Oct 12 '15 at 16:08

Alternative method with tikz-cd: crossing over is easy, it's just necessary at times to draw an arrow later.

For the Y node, some adjustment is necessary for getting parallel arrows.

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

\begin{document}

\begin{tikzcd}[column sep=1.5em]
\{\iota\} \sqcup A^0
  \arrow[dd,swap,"\bigsqcup\limits_{\alpha\in A^0} g_\alpha"]
%  \arrow[rddd]% later, for crossing over
  \arrow[rr] &&
\{\iota\} \sqcup A
  \arrow[dd]
%  \arrow[rddd,"\rho^{\{\iota\}}(f)"]% later, for crossing over
  \arrow[rr,"\phi"] &&
X
  \arrow[rrddd,xshift=-.4em,end anchor={[xshift=.4em]north},"g"] \\ \\
\bigsqcup\limits_{\alpha\in A^0} P_\alpha
  \arrow[rddd,swap,"\subset"]
  \arrow[rr] &&
\widehat{A}
  \arrow[rddd,dashed,swap,pos=0.25,"\exists!"]
  \arrow[rruu,dashed,"\exists!"] \\
&\{\iota\} \sqcup B^0
  \arrow[luuu,<-,crossing over]% reversed!
  \arrow[dd,"\bigsqcup\limits_{\beta\in B^0} g_\beta"]
  \arrow[rr,crossing over] &&
\{\iota\} \sqcup B
  \arrow[luuu,<-,crossing over,swap,pos=0.3,"\rho^{\{\iota\}}(f)"]% reversed!
  \arrow[rrr,"\psi"]
  \arrow[dd] &&&
Y \\ \\
&\bigsqcup\limits_{\beta\in B^0} P_\beta
  \arrow[rr] &&
\widehat{B}
  \arrow[rrruu,dashed,"\exists!"]
\end{tikzcd}

\end{document}

score 2 · Answer 3 · answered Nov 16 '15 at 16:12

In addition to egreg's solution, here is a not hard coded correction in order to get parallel arrows. I also changed a bit the way to annotate this just in order to show possibilities. Choose the way you like.

% arara: pdflatex

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

\begin{document}
\[
\begin{tikzcd}[column sep={15ex,between origins}, row sep={5em,between origins}]
    \{\iota\} \sqcup A^0 \arrow{d}[swap]{\bigsqcup\limits_{\alpha\in A^0} g_\alpha} \arrow{rr} 
    & &[-2.5em]
    \{\iota\} \sqcup A \arrow{d}\arrow{rr}{\phi} 
    & &[-5em]
    X \arrow{rdd}{g} 
    \\
    \bigsqcup\limits_{\alpha\in A^0} P_\alpha \arrow{rdd}[swap]{\subset}\arrow{rr} 
    & & 
    \widehat{A} \arrow[dashed]{rdd}[swap,pos=.25]{\exists!}\arrow[dashed]{rru}{\exists!} 
    \\
    &
    \{\iota\} \sqcup B^0 \arrow[leftarrow,crossing over]{luu}\arrow{d}{\bigsqcup\limits_{\beta\in B^0} g_\beta}\arrow[crossing over]{rr}
    & &
    \{\iota\} \sqcup B \arrow[leftarrow,crossing over]{luu}[swap,pos=0.3]{\rho^{\{\iota\}}(f)}\arrow{rr}{\psi}\arrow{d}
    & &
    Y 
    \\
    &
    \bigsqcup\limits_{\beta\in B^0} P_\beta \arrow{rr} 
    & &
    \widehat{B} \arrow[dashed]{rru}{\exists!}
    & &
\end{tikzcd}
\]
\end{document}

I keep forgetting between origins! – egreg Nov 16 '15 at 21:52 — egreg, Nov 16 '15 at 21:52

Putting a hole at the right position (xypics)

3 Answers3