I've been looking for arrow symbols that form a right angle, and 45' angles pointing to the upper-left and left of the symbol (combination of \leftarrow and \nwarrow, essentially). Generally, all combinations of the three arrows \leftarrow, \uparrow and \nwarrow.
I tried \mathrlap...
$\mathrlap{\leftarrow}\uparrow$ but the results were less than satisfactory --the origin of the arrows should be the same point.
Any advice would be appreciated.
The following example uses \uparrow as base unit, because the arrow reaches from the bottom to the top of the bounding box without interfering side bearings.
Macro \@leftuparrow takes three arguments with values 0 and 1 each to denote the left, northwest, and up arrow component of the total symbol.
As the examples show, a font for the up arrow should be used, where the tip is not too broad.
\documentclass{article} \usepackage{amsmath} %\usepackage{MnSymbol} %\usepackage{mathabx} \usepackage{graphicx} \makeatletter \newcommand*{\leftuparrow}{% \@leftuparrow101% } \newcommand*{\loweredleftarrow}{\@leftuparrow100} \newcommand*{\leftnwarrow}{\@leftuparrow110} \newcommand*{\leftnwuparrow}{\@leftuparrow111} \newcommand*{\nwuparrow}{\@leftuparrow011} \newcommand*{\@leftuparrow}[3]{% \mathrel{% \mathpalette{\@@leftuparrow{#1}{#2}{#3}}{}% }% } \newcommand*{\@@leftuparrow}[5]{% % #1: left % #2: nw % #3: up % #4: math style % #5: unused \sbox0{$#4\uparrow\m@th$}% base unit % % Correction of depth to compensate line width \sbox6{$#4\mkern.25mu\m@th$}% \dimen@=\dp0 % \advance\dimen@ by -\wd6 % \dp0=\dimen@ % \sbox2{% \rotatebox[origin=b]{45}{\hbox to\z@{\hss\copy0\hss}}% }% \sbox4{% \rotatebox[origin=b]{90}{\copy0}% }% \dimen@=\z@ \dimen6=\z@ \ifnum#1#2>\z@ \kern4\wd6 % left side bearing \fi \ifnum#1=1 % leftarrow \copy4 % \dimen@=\wd4 % \dimen6=\wd6 % \fi \ifnum#2=1 % nwarrow \ifdim\dimen@>\z@ \llap{\copy2}% \else \copy2 % \dimen@=\wd2 % \dimen6=.7071\wd6 % \fi \fi \ifnum#3=1 % uparrow \ifdim\dimen@>\z@ \kern-.5\wd0 % \fi {\uparrow}% \else \kern4\wd6 % right side bearing \fi \kern\dimen6 % } \makeatother \begin{document} \newcommand*{\test}{% \loweredleftarrow \leftnwarrow \leftnwuparrow \leftuparrow \nwuparrow \uparrow } \setlength{\fboxsep}{0pt}% \setlength{\fboxrule}{.1pt}% \begin{gather*} \fbox{$\leftarrow$} \fbox{$\uparrow$} \fbox{$\nwarrow$} \\ \fbox{$\loweredleftarrow$} \fbox{$\leftnwarrow$} \fbox{$\leftnwuparrow$} \fbox{$\leftuparrow$} \fbox{$\nwuparrow$} \\ \test\\ \scriptstyle\test\\ \scriptscriptstyle\test \end{gather*} \end{document} Result for Computer Modern fonts:
Result for MnSymbol:
Result for mathabx:
\documentclass{article} \usepackage{mathtools} \begin{document} \[ \raisebox{-1ex}{$\leftarrow$}\mathllap{\nwarrow}\mkern-10mu\uparrow \] \end{document} 
% arara: pdflatex \documentclass{article} \usepackage{tikz-cd} \newcommand{\myArrow}[1]{\arrow[start anchor=center, end anchor=center]{#1}} \newcommand{\nothing}{\phantom{\bullet}} \begin{document} \begin{tikzcd}[row sep=.5cm, column sep= .5cm] \nothing&\nothing \\ \nothing&\nothing\myArrow{l}\myArrow{u} \end{tikzcd} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{tikzcd}[row sep=.5cm, column sep= .5cm] \nothing&\nothing \\ \nothing&\nothing\myArrow{ul}\myArrow{l} \end{tikzcd} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{tikzcd}[row sep=.5cm, column sep= .5cm] \nothing&\nothing \\ \nothing&\nothing\myArrow{ul}\myArrow{u}\myArrow{l} \end{tikzcd} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{tikzcd}[row sep=.5cm, column sep= .5cm] \nothing&\nothing \\ \nothing&\nothing\myArrow{ul}\myArrow{u} \end{tikzcd} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{tikzcd}[row sep=.5cm, column sep= .5cm] \nothing&\nothing \\ \nothing&\nothing\myArrow{l} \end{tikzcd} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{tikzcd}[row sep=.5cm, column sep= .5cm] \nothing&\nothing \\ \nothing&\nothing\myArrow{u} \end{tikzcd} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{tikzcd}[row sep=.5cm, column sep= .5cm] \nothing&\nothing \\ \nothing&\nothing\myArrow{ul} \end{tikzcd} \end{document} 
tikz is available it is easier to do \draw[<->] (-.5cm,0)--(0,0)--(0,.5cm);