The spacing of `\to`

Question

Let's say I use \def\searrow{\mathbin{\text{\rotatebox[origin=c]{-33}{$\to$}}}} instead of the \searrow.

But \mathbin doesn't seem to provide the spacing that \to has; what should it be? Other suggestions are welcome.

According to The LaTeX Companion, arrow symbols (including \rightarrow, aka \to) are of type mathrel, not mathbin. — Mico, Jan 16 '14 at 15:51
Relevant: What is the difference between \mathbin vs. \mathrel? — Werner, Jan 16 '14 at 16:08
Thank you; yet, to my eye it looks worse with mathrel (too much spacing). In what way might the other ingredients be influencing the spacing in this case? The thing is that the \to arrow seems to be smaller after rotation. — user66081, Jan 16 '14 at 16:14
Any particular reason why you are doing your own version of \searrow? — daleif, Jan 16 '14 at 16:26
What matters for total spacing, i.e., the visual width of the symbol plus the space on either side, are the dimensions of the "box" (rectangle) that contains the symbol. After the box has been rotated by 30 degrees or so, its total left-to-right (horizontal) width exceeds that of the unrotated \to symbol. Hence, a moderately-rotated \to symbol looks like it has more whitespace on either side than the unrotated symbol. By the way, echoing @daleif's comment, is there a reason why you're not using the predefined \searrow macro? — Mico, Jan 16 '14 at 16:38

score 2 · Accepted Answer · answered Feb 04 '17 at 21:52

You have to do a small backspacing and then leave a small space after the arrow.

\documentclass{article}
\usepackage{amsmath,graphicx}

\renewcommand\searrow{%
  \mathrel{\mspace{-1.5mu}}%
  \mathrel{\text{\rotatebox[origin=c]{-33}{$\to$}}}%
  \mathrel{\mspace{.808mu}}
}

\begin{document}

$A\to B$

$A\searrow B$

\end{document}

The value of -1.5mu has been evaluated by eye; the forward spacing instead is made so that the space occupied is the same up to the third decimal in points.

The spacing of `\to`

1 Answers1