tikz and limits
the same without tikZ
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\varinjlim\limits_{\mathcal{I}}
\left(\begin{array}{l}
A\to B \\
\,\downarrow \\
C
\end{array}\right) = C \underset{A}{\amalg} B
\]
\end{document}
You have to set the baseline of the tikzpicture
to the vertical center of it; also it's better to use left delimiter
and right delimiter
rather than \left
and \right
:
\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}
\[
\varinjlim\limits_{\mathcal{I}}
\begin{tikzpicture}[baseline=(current bounding box.center)]
\matrix (m) [
matrix of math nodes,
row sep=1em,
column sep=1em,
left delimiter=(,
right delimiter=),
]{
A & B \\
C \\
};
\path[-stealth]
(m-1-1) edge node [left] {} (m-2-1)
edge node [above] {} (m-1-2);
\end{tikzpicture}
= C \underset{A}{\amalg} B
\]
\end{document}
I changed \cI
into \mathcal{I}
, just a guess; also \coprod
should not be used as a binary operator. If you really want to set the subscript underneath \amalg
use \underset
.
With the corrections suggested by Qrrbrbirlbel:
\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}
\[
\varinjlim\limits_{\mathcal{I}}
\begin{tikzpicture}[baseline=-\the\dimexpr\fontdimen22\textfont2]
\matrix (m) [
matrix of math nodes,
row sep=1em,
column sep=1em,
outer sep=0pt,inner sep=0pt,
nodes={inner sep=.3333em},
left delimiter=(,
right delimiter=),
]{
A & B \\
C \\
};
\path[-stealth]
(m-1-1) edge node [left] {} (m-2-1)
edge node [above] {} (m-1-2);
\end{tikzpicture}
= C \underset{A}{\amalg} B
\]
\end{document}