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}