# What is the correspondence symbol with a hat or wedge over equals?

I would like to produce a new equal symbol looking like this. .

What is the best way to do that ? Pure TeX ? TiKz ?

I'm looking for ways to do that rather than to a ready to use solution such as to be able to other new symbols alone.

• This looks to me like Unicode U+2259, estimates (\wedgeq in the stix fonts). What is your intended meaning? (If it's the same. I'll try to add an answer, but since I don't have easy access to a tex system at the moment, including a visual will be difficult.) – barbara beeton Sep 28 '19 at 1:24

Before somebody suggests MnSymbol

\documentclass{article}
\usepackage{amsmath}

\makeatletter
\providecommand{\hateq}{\mathrel{\mathpalette\my@hat@eq\relax}}
\newcommand{\my@hat@eq}[2]{%
\begingroup
\sbox\z@{$\m@th#1=$}%
\ooalign{%
\hidewidth\raisebox{-0.3\ht\z@}{$\m@th#1\widehat{}$}\hidewidth\cr
\box\z@\cr
}%
\endgroup
}
\makeatother

\begin{document}

$A\cong B\hateq C$

\end{document}


With \hat instead of \widehat:

\documentclass{article}
\usepackage{amsmath}

\makeatletter
\providecommand{\hateq}{\mathrel{\mathpalette\my@hat@eq\relax}}
\newcommand{\my@hat@eq}[2]{%
\begingroup
\sbox\z@{$\m@th#1=$}%
\ooalign{%
\hidewidth\raisebox{-0.2\ht\z@}{$\m@th#1\hat{}$}\hidewidth\cr
\box\z@\cr
}%
\endgroup
}
\makeatother

\begin{document}

$A\cong B\hateq C$

\end{document}


• @campa Yes, thanks! – egreg Sep 27 '19 at 15:07
• How is this different from U+2259, estimates (\wedgeq in the stix fonts)? (I know the shape is slightly different, but what about the meaning?) – barbara beeton Sep 27 '19 at 18:24
• @barbarabeeton It's the same, I guess. No idea what meaning the OP is assigning to the symbol. – egreg Sep 27 '19 at 20:29
• @projetmbc -- I don't know of any equivalent of detexify for Unicode, but it might be reasonable to make a suggestion to the detexify developer to add Unicode IDs to the returned information. It doesn't look like the stix font info has gotten into detexify, so that's also a plausible request. (The stix font and the addition of many symbols to unicode progressed in tandem.) – barbara beeton Sep 28 '19 at 14:19
• @barbarabeeton maybe shapecatcher.com? – Rmano Sep 29 '19 at 10:48

What you want could be obtained using

\documentclass[12pt]{article}
\usepackage{amsmath}
\newcommand{\circumeq}{\mathrel{\widehat{=}}}
\begin{document}
\begin{equation*}
a \circumeq b
\end{equation*}
\end{document}


• Better \mathrel{\widehat{=}} – egreg Sep 27 '19 at 14:38
• Thanks but how to do that "all by hand" if I want to do other new symbols using existing symbols and possibly simple polylines. – projetmbc Sep 27 '19 at 14:39
• @projetmbc: An example of how to make symbols in TixZ is at How can I get a big cross to denote a generalized cartesian product?. – Peter Grill Sep 27 '19 at 14:43
• @projetmbc OK. I suggest then having a look at (tex.stackexchange.com/questions/22371/…) with a fine course on \ooalign – Denis Sep 27 '19 at 14:47
• Another possibility is playing with the \stackinset command from  stackengine (see § 2.3.10 in the documentation). – Bernard Sep 27 '19 at 18:09

It seems like a variation of the congruence symbol. Denis's answer is aligned with the =. To align it with \cong you need more trickery:

\documentclass{article}

\usepackage{amsmath} % necessary for correct scaling of \widehat

\makeatletter
\newcommand*{\varcong}{\mathrel{\mathpalette\@varcong\relax}}
\newcommand*{\@varcong}[2]{\vcenter{\hbox{\m@th$#1\widehat{\mathop=}$}}}
\makeatother

\begin{document}

$a \cong a \varcong a = a \mathrel{\widehat{=}} a$

$\scriptstyle a \cong a \varcong a = a \mathrel{\widehat{=}} a$

$\scriptscriptstyle a \cong a \varcong a = a \mathrel{\widehat{=}} a$

\end{document}


• This symbol is used by the B language to specify and prove programs. – projetmbc Sep 27 '19 at 15:22

The ≙ symbol is U+2259 in Unicode. It has the name \wedgeq in several newer packages, including unicode-math and stix. It is \hateq in two older ones, mnsymbol and fdsymbol. One completely obsolete one, boisik, has \corresponds.

Edit: Sorry, just noticed that Barbara Beeton said that in a comment first. Props!