matrix with tikz

Question

I would like to reproduce the diagram in 2.1 of https://www3.nd.edu/~wgd/Dvi/CalculatingSimplicialLocalizations.pdf

So far I have this piece of code but I don't know how to fill cells with horizontal lines as in the link above:

\begin{tikzpicture}
\matrix[matrix of math nodes]{

  & C_{0,1} & C_{0,2} &    &  \cdots       &        & C_{0,n-1} &   \\

  &         &         &    &               &        &           &   \\

X &         &         &    &               &        &           & Y \\

  &         &         &    &               &        &           &   \\

  &         &         &    &               &          &         &   \\

};
\end{tikzpicture}

Answer 1

This can be done by tikz-cd, xy or pure pstricks and tikz.

I introduced some dimensions (e.g. [-2\jot]). You may want to modify or delete them until the diagram fits your desired style.

Here you are:

% arara: pdflatex

\documentclass{article}
\usepackage{mathtools}
\usepackage{tikz-cd}
\makeatletter
\DeclareRobustCommand{\rvdots}{%
    \vbox{
        \baselineskip4\p@\lineskiplimit\z@
        \kern-\p@
        \hbox{.}\hbox{.}\hbox{.}
    }}
    \makeatother

\begin{document}
\[
\begin{tikzcd}
& C_{0,1}\arrow[dash]{r}\arrow{d} & C_{0,2}\arrow[dash]{r}\arrow{d} &[-2\jot] \cdots\arrow[dash]{r} &[-2\jot] C_{0,n-1}\arrow{d} & \\
& C_{1,1}\arrow[dash]{r}\arrow{dd} & C_{1,2}\arrow[dash]{r}\arrow{dd} & \cdots\arrow[dash]{r} & C_{1,n-1}\arrow{dd} & \\
X \arrow[dash]{uur}\arrow[dash]{ur}\arrow[dash]{ddr} &&&&& Y\arrow[dash]{uul}\arrow[dash]{ul}\arrow[dash]{ddl} \\[-7.2\jot]
& \rvdots \arrow{d} & \rvdots \arrow{d} && \rvdots \arrow{d} & \\
& C_{k,1}\arrow[dash]{r} & C_{k,2}\arrow[dash]{r} & \cdots\arrow[dash]{r} & C_{k,n-1} &
\end{tikzcd}
\tag{*}\]
\end{document}

Answer 2

Similar to LaRiFaRi, but with less dramatic length for the vertical arrows to the dots. The trick is to use mock rows and shortening row sep.

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

\newcommand{\cvdots}{\raisebox{-.4ex}[0pt][0pt]{$\vdots$}\mathstrut}

\begin{document}

\[
\begin{tikzcd}[row sep=1em]
 & C_{0,1} \arrow[dddl,dash] \arrow[r,dash] \arrow[dd]
 & C_{0,2} \arrow[r,dash] \arrow[dd]
 & \cdots \arrow[r,dash]
 & C_{0,n-1} \arrow[dddr,dash] \arrow[dd]
\\ \\
 & C_{1,1} \arrow[dl,dash] \arrow[r,dash] \arrow[dd]
 & C_{1,2} \arrow[r,dash] \arrow[dd]
 & \cdots \arrow[r,dash]
 & C_{1,n-1} \arrow[dr,dash] \arrow[dd]
\\
X \arrow[dddr,dash] & & & & & Y \arrow[dddl,dash]
\\
& \cvdots \arrow[dd] & \cvdots \arrow[dd] & & \cvdots \arrow[dd]
\\ \\
 & C_{k,1} \arrow[r,dash]
 & C_{k,2} \arrow[r,dash]
 & \cdots \arrow[r,dash]
 & C_{k,n-1}
\end{tikzcd}
\]

\end{document}

Answer 3

To make lines with a matrix as base you will need to name the matrix and use \path. I removed your empty columns and used this example as reference: http://www.texample.net/tikz/examples/commutative-diagram-tikz/

To make the vertical dots I create named nodes in the middle of the edges that go from the second row to the last row. I fill each node with white to hide the edge and add another edge from the second row to the node to get the arrow.

Unfortunately \vdots has extra space on top, so instead I used the alternative presented here: How to vertically center the \vdots in this node?

The forth row is empty and has no height. To make sure X and Y appear centered you need to add some phantom content to the row, like \phantom{C}.

% vertical dots without extra space on top
\makeatletter
\DeclareRobustCommand{\rvdots}{%
  \vbox{
    \baselineskip4\p@\lineskiplimit\z@
    \kern-\p@
    \hbox{.}\hbox{.}\hbox{.}
  }}
\makeatother

\begin{tikzpicture}
\matrix (m) [matrix of math nodes,row sep=2em,column sep=2em]{
  & C_{0,1} & C_{0,2} & \cdots & C_{0,n-1}   &   \\
  & C_{1,1} & C_{1,2} & \cdots & C_{1,n-1}   &   \\
X &         &         &        &             & Y \\
  &         &         &        & \phantom{C} &   \\
  & C_{k,1} & C_{k,2} & \cdots & C_{k,n-1}   &   \\
};
\path[-stealth]
    % X
    (m-3-1) edge [-] (m-1-2)
            edge [-] (m-2-2)
            edge [-] (m-5-2)
    % first row
    (m-1-2) edge [-] (m-1-3)
            edge (m-2-2)
    (m-1-3) edge [-] (m-1-4)
            edge (m-2-3)
    (m-1-4) edge [-] (m-1-5)
    (m-1-5) edge [-] (m-3-6)
            edge (m-2-5)
    % second row
    (m-2-2) edge [-] (m-2-3)
            edge node(vdots-2) [fill=white] {$\rvdots$} (m-5-2)
            edge (vdots-2)
    (m-2-3) edge [-] (m-2-4)
            edge node(vdots-3) [fill=white] {$\rvdots$} (m-5-3)
            edge (vdots-3)
    (m-2-4) edge [-] (m-2-5)
    (m-2-5) edge [-] (m-3-6)
            edge node(vdots-5) [fill=white] {$\rvdots$} (m-5-5)
            edge (vdots-5)
    % last row
    (m-5-2) edge [-] (m-5-3)
    (m-5-3) edge [-] (m-5-4)
    (m-5-4) edge [-] (m-5-5)
    (m-5-5) edge [-] (m-3-6)
    % Y is (m-3-6)
;
\end{tikzpicture}

Answer 4

A bit unfriendly in places, but it pretty much does the job...

\documentclass[tikz,border=5pt]{standalone}
\begin{document} 
\begin{tikzpicture}[x=1.5cm,y=1.25cm,>=stealth]
\foreach \i [count=\y] in {0,1,\vdots,k}
  \foreach \j [count=\x] in {1,2,\ldots,n-1}
    \node (C-\y-\x) at (\x,-\y+2.5) 
      {$\ifnum\y=3\ifnum\x=3\else\i\fi\else\ifnum\x=3\j\else C_{\i,\j}\fi\fi$};
\node at (0,0) (X) {$X$};
\node at (5,0) (Y) {$Y$};    
\foreach \j in {1,2,4}{
  \foreach \i [count=\k from 2] in {1,2,3}
     \draw [->] (C-\j-\i) -- (C-\j-\k) (C-\i-\j) -- (C-\k-\j);
  \draw (X) -- (C-\j-1) (C-\j-4) -- (Y);
}
\end{tikzpicture}
\end{document}