Pushouts and pullbacks

Question

How do I typeset that little symbol which indicates that a commutative square is a pushout or pullback square? I usually draw my diagrams in XY-pic, so I'd be most interested in how to do it there.

Added: The desired effect is a diagram like

Edit: The problem is in typesetting the actual symbol, the XY-pic user's guide says how to place it.

Answer 1

Put these guys in your preamble:

\newcommand{\po}{\ar@{}[dr]|{\text{\pigpenfont R}}}
\newcommand{\pb}{\ar@{}[dr]|{\text{\pigpenfont J}}}

This uses the package pigpen for the symbols in the middle. The \ar@{} specifies an invisible arrow, and |{-} smacks some text in the middle of it. Whenever you want a pushout (or pullback) you put \po (or \pb) in the upper left of your square. Here is some sample code:

\xymatrix{X\ar[r]\ar[d]_{i_0}\po &\ast\ar[d]\\
    X\times I\ar[r]&CX}

and here is the picture:

Answer 2

I'm posting this more so that people know what K.J. Moi is talking about! Here's how to do it in TikZ:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}

\begin{tikzpicture}[>=latex]
\node (w) at (0,0) {\(W\)};
\node (x) at (0,-2) {\(X\)};
\node (y) at (2,0) {\(Y\)};
\node (z) at (2,-2) {\(Z\)};
\draw[->] (w) -- (y);
\draw[->] (w) -- (x);
\draw[->] (x) -- (z);
\draw[->] (y) -- (z);
\begin{scope}[shift=($(w)!.5!(z)$)]
\draw +(-.5,0) -- +(0,0)  -- +(0,.5);
\fill +(-.25,.25) circle (.05);
\end{scope}
\end{tikzpicture}
\end{document}

And here's the output:

(pdf, and this is not to be confused with these instructions for opening a door)

So strictly speaking, it's not a symbol itself but a couple of lines set into the diagram. I don't know how to do this in XY as I'm a die-hard practitioner of TikZ, but hopefully seeing the picture will help others figure it out for you!

(Note that there are probably slicker ways of doing it in TikZ as well, I was going for speed here as I guessed not everyone would know what a pullback diagram is!)

(Edited to add the dot and to shift the symbol a bit closer towards the Z; I'm not sure exactly how far it should be shifted, but the .5 in the expression ($(w)!.5!(z)$) controls how far between the W and Z the point of the symbol lies; bigger numbers shift it to towards Z.)

Answer 3

The following solution is from the tikzcd documentation, Version 0.9e October 30, 2014:

\begin{tikzcd}
    A \arrow[r] \arrow[d]
    \arrow[dr, phantom, "\ulcorner", very near start]
    & B \arrow[d] \\
    C \arrow[r]
    & D
\end{tikzcd}

You can change the corner to \lrcorner.

Answer 4

Use the Comprehensive LaTeX Symbol List to find it or use the Detexify tool.

You may use the symbols \lrcorner, \ulcorner, \llcorner or \urcorner of the amssymb package.

A dot might be put into the corner for instance by $\rlap{$\cdot$}\lrcorner$. Or use \mathrlap as suggested in the comments.

Here's a possible definition for that symbol:

\newcommand*{\corner}{\mbox{\LARGE{$\mathrlap{\cdot}\lrcorner$}}}

Insert small spacing if you would like to adjust the position of the dot within the corner symbol.

Here's a complete minimal example producing a complete matrix:

\documentclass{article}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage{xypic}
\newcommand*{\corner}{\mbox{\LARGE{$\mathrlap{\cdot}\lrcorner$}}}
\begin{document}
\xymatrix{%
  \ar @{} [dr] | \corner A \ar[r] \ar[d] &B \ar[d] \\
  C \ar[r] & D
}
\end{document}

Answer 5

Here is a solution using pic of TikZ.

\documentclass[tikz,border=5mm]{standalone}
\begin{document}
\begin{tikzpicture}[>=latex,bmrd/.pic={
\draw (2mm,0)--++(-90:2mm)--++(180:2mm);
\fill (1mm,-1mm) circle(1pt);
}]
\def\a{1.5}
\path
(0,0)        node (W) {$W$}
+(-45:.5*\a) pic[scale=1.2,red]{bmrd}
+(0:\a)      node (Y) {$Y$}
++(-90:\a)   node (X) {$X$}
+(0:\a)      node (Z) {$Z$};
\draw[->] (W)--(Y);
\draw[->] (W)--(X);
\draw[->] (X)--(Z);
\draw[->] (Y)--(Z);
\end{tikzpicture}
\end{document}