# problem making a commutative diagram in tikz using positioning library

I am using the positioning TikZ library to make a commutative diagram. I am having trouble getting the bottom arrow to be horizontal.

Here is the output with my problem: the positioning library is on the left, and the output I thought I should get (done with the matrix library) is on the right. The problem is the arrow above $\psi^{-1}$: it is not exactly horizontal on the left version of the diagram.

What am I doing wrong? Is there a reason I shouldn't be getting the same picture on the left and on the right?

And here is my code:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{matrix,positioning}

\begin{document}

$\begin{tikzpicture}[ baseline=(current bounding box.center)] % Tell it where the nodes are \node (F) {M^P \cap \pi_W^{-1}( U) }; \node (E) [right=of F] {V}; \draw[->] (F)-- node [above] {\footnotesize \pi_V} (E); \node (C) [below=of F] {U \subset W}; \draw[->] (F)-- node [left] {\footnotesize \pi_W} (C); \node (B) [below=of E] {I \times V}; \draw[->] (C)-- node [below] {\footnotesize \psi^{-1}} (B); \draw[<-] (E)-- (B); \end{tikzpicture} = \begin{tikzpicture}[every node/.style={midway}, baseline=(current bounding box.center)] \matrix[column sep={6em,between origins}, row sep={3em}] at (0,0) { \node (F) {M^P \cap \pi_W^{-1}( U) }; & \node (E) {V};\\ \node (C) {U \subset W};& \node (B) {I \times V};\\}; \draw[->] (F)-- node [above] {\footnotesize \pi_V} (E); \draw[->] (F)-- node [left] {\footnotesize \pi_W} (C); \draw[->] (C)-- node [below] {\footnotesize \psi^{-1}} (B); \draw[<-] (E)-- (B); \end{tikzpicture}$
\end{document}

• Take a look at the snake lemma (tex.stackexchange.com/a/3894/86), the bit starting "(Added in edit)". Mar 19, 2012 at 13:52
• Though looking in more detail at your code, that's the explanation but the solutions are different. Mar 19, 2012 at 13:54
• Not directly related to the question itself, but there is a package for creating commutative diagrams that uses TikZ, tikz-cd. Mar 19, 2012 at 14:02
• The top left node has a subscript, hence it has a greater depth. The below=of x on both nodes simply positions the new node some standard distance below x (node distance actually, which is 1cm and 1cm by default). below uses the south anchor, so if one of the nodes has a greater depth it will also have a slightly lower anchor (since the centers are aligned, due to the placement using right=of...). This causes the new nodes to not be aligned horizontally. You can remedy this by using the center anchor explicitly. when saying below=of .... Mar 19, 2012 at 14:28

This is normal. The node of the first lines are different heights. You need to use option on grid

You can remove every node/.style={midway} in the second picture

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{matrix,positioning}

\begin{document}

\begin{tikzpicture}[every node/.style={on grid},  baseline=(current bounding box.center),node distance=2]
% Tell it where the nodes are
\node (F) {$M^P \cap \pi_W^{-1}( U)$};
\node (E) [right=of F] {$V$};
\draw[->] (F)-- node [above] {\footnotesize $\pi_V$} (E);
\node (C) [below=of F] {$U \subset W$};
\draw[->] (F)-- node [left] {\footnotesize $\pi_W$} (C);
\node (B) [below=of E] {$I \times V$};
\draw[->] (C)-- node [below] {\footnotesize $\psi^{-1}$} (B);
\draw[<-] (E)-- (B);
\end{tikzpicture}
=
\begin{tikzpicture}[  baseline=(current bounding box.center)]
\matrix[column sep={6em,between origins},
row sep={3em}] at (0,0)
{
\node (F) {$M^P \cap \pi_W^{-1}( U)$}; &
\node (E) {$V$};\\
\node (C) {$U \subset W$};&
\node (B) {$I \times V$};\\};
\draw[->] (F)-- node [above] {\footnotesize $\pi_V$} (E);
\draw[->] (F)-- node [left] {\footnotesize $\pi_W$} (C);
\draw[->] (C)-- node [below] {\footnotesize $\psi^{-1}$} (B);
\draw[<-] (E)-- (B);
\end{tikzpicture}

\end{document}