# Widehat with right arrow

I am searching for a wide hat with a right arrow. So far I've found only the regular wide hat (\widehat, shown at the left of the image), or a curved right arrow (\curvearrowright).

Is there a way to get something like what I have badly drawn at the right of the image?

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Welcome to TeX.sx!. I have included your image. Soon, you will have enough reputation to do it yourself. –  Qrrbrbirlbel Oct 7 '12 at 17:59
Thanks for including the image! –  Paul22 Oct 7 '12 at 18:16

I provide two TikZ solutions.

1. The first one uses uses an empty node before and after the formula that belongs under the wide hat.
• Disadvantage: The added vertical height will not be taken into calculation
• Advantage: Math style (display, text, script, scriptscript) is not effected (other then in the \vphantom.
2. The second one is a node.
• Disadvantage: Math style is lost. (Change the first \frac into a \tfrac to realize.)
• Advantage: Vertical height is known to math mode.

Another disadvantage over \widehat is, that the line width doesn't change over the course of the path.

## MWE

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc,arrows}

% Solution 1
\newcounter{tikzmark}
\newcommand{\tikzmark}[2][]{\tikz[baseline, inner sep=0, overlay, remember picture] \node[anchor=base] (#2) {\vphantom{#1}};}

\newcommand{\widehatarrow}[1]{%
\tikzmark[#1]{widehatarrow\thetikzmark} #1 \tikzmark[#1]{endwidehatarrow\thetikzmark}%
\tikz[overlay, remember picture] \draw[-to] (widehatarrow\thetikzmark.north)++(.05em,.4ex) -- ($(widehatarrow\thetikzmark.north)!.5!(endwidehatarrow\thetikzmark.north)+(.05em,.8ex)$) -- ($(endwidehatarrow\thetikzmark.north)+(.05em,.4ex)$);%
\stepcounter{tikzmark}%
}

% Solution 2
\newcommand{\Widehatarrow}[1]{%
\tikz[baseline] {
\node[inner sep=0, anchor=base] (widehatarrow){$#1$};
\draw[-to] (widehatarrow.north west)++(.05em,.4ex) -- ($(widehatarrow.north west)!.5!(widehatarrow.north east)+(.05em,.8ex)$) -- ($(widehatarrow.north east)+(.05em,.4ex)$);%
}%
}

\begin{document}
$$% Solution 1 \widehatarrow{A, B} \cdot \widehatarrow{x, y} = \widehatarrow{A, y, B, x}$$

$$% Solution 2 \Widehatarrow{A, B} \cdot \Widehatarrow{x, y} = \Widehatarrow{A, y, B, x}$$

$$% Compare both solutions \frac{1}{\widehatarrow{A, B} \cdot \Widehatarrow{A, B}} = \frac{1}{\widehatarrow{A, B} \cdot \widehatarrow{A, B}} = \frac{1}{\Widehatarrow{A, B} \cdot \Widehatarrow{A, B}}$$
\end{document}


## Output

(By egreg) It's possible to overcome the limitations with \Widehatarrow, by using a \vphantom and \mathpalette:

\newcommand{\Widehatarrow}[1]{\mathpalette\Widehatarrowaux{#1}}
\newcommand{\Widehatarrowaux}[2]{%
\vphantom{#1\widehat{#2}}
\tikz[baseline] {
\node[inner sep=0, anchor=base] (widehatarrow){$#1#2$};
\draw[-to] (widehatarrow.north west)++(.05em,.4ex)
-- ($(widehatarrow.north west)!.5!(widehatarrow.north east)+(.05em,.8ex)$)
-- ($(widehatarrow.north east)+(.05em,.4ex)$);%
}%
}


Now the height of \Widehatarrow{AB} will be the same as the height of \widehat{AB} and the sizes will be respected, as shown by trying

$\Widehatarrow{AB}=A_{\Widehatarrow{XY}}$

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Anyone, who has an idea how to solve the problems (“disadvantage”), is welcome to comment or point me in the right direction. :) –  Qrrbrbirlbel Oct 7 '12 at 18:55
Both problems can be solved with some cognoscenti trickery, for \Widehatarrow. I'll add it to your solution. –  egreg Oct 7 '12 at 19:11
The first solution is quite buggy with me (see mirari.fr/BJr2 ) (and it compiles differently every time). Anyway, the second (and its modified version) works perfectly. Thanks! –  Paul22 Oct 7 '12 at 20:04
@Paul22 Yes, the first one needs two compilations and is superseded by the second one (especially with egreg's addition), anyway. –  Qrrbrbirlbel Oct 7 '12 at 20:24