How to put arrows pointing to equalities to explain each step

Question

What´s the best way to explain every equality? With arrows like in

Anything similar to this

Answer 1

I can suggest three possibilities.

\documentclass{article}
\usepackage{amsmath}

% for the first solution
\usepackage{mathtools}
\newcommand{\equalexpl}[1]{%
  \underset{\substack{\uparrow\\\mathrlap{\text{\hspace{-1em}#1}}}}{=}}


\begin{document}

\begin{align*}
f(ax+by) 
&\equalexpl{because $f(u+v)=f(u)+f(v)$}
  f(ax) + f(by) \\
&\equalexpl{because $f(cu)=cf(u)$}
 af(x) + bf(y)\\
\end{align*}

\begin{alignat*}{2}
f(ax+by)
&= f(ax) + f(by)\qquad &&\text{because $f(u+v)=f(u)+f(v)$} \\
&= af(x) + bf(y)\qquad &&\text{because $f(cu)=cf(u)$}
\end{alignat*}

\begin{align*}
&f(ax+by)\overset{(1)}{=} f(ax) + f(by) \overset{(2)}{=} af(x) + bf(y)\\[2ex]
&\text{(1) because $f(u+v)=f(u)+f(v)$} \\
&\text{(2) because $f(cu)=cf(u)$}
\end{align*}

\end{document}

Answer 2

Here's a solution using TikZ:

\documentclass{article}
\usepackage{tikz}

\newcounter{mybox}
\newcommand\tikzmark[1]{%
\tikz[remember picture,overlay] \node[inner xsep=0pt] (#1) {};
}
\newcommand\ColorBox[2][]{%
\stepcounter{mybox}%
\node[draw=red!70!black,fill=red!20,align=left,#1] (box\themybox) {#2};
}

\begin{document}

\[
a(b + c) \tikzmark{a}= ab+ac \tikzmark{b}= ba+ca.
\]
\begin{tikzpicture}[remember picture,overlay]
\ColorBox[xshift=3cm]{by distributivity}
\draw[red!70!black,->] (box1) -| ([xshift=7.5pt]a.south west);
\ColorBox[xshift=10cm,fill=blue!30,draw=blue]{by commutativity}
\draw[blue,->] (box2) -| ([xshift=7.5pt]b.south west);
\end{tikzpicture}

\end{document}

\ColorBox has one optional argument to control options for the \node used to draw the box, and a mandatory argument for the explanatory text. The associated node is automatically named box<number>. Using \tikzmark you mark any element that will be the end of the arrow and then you can draw the arrow with a standard \draw command.

Answer 3

If I understand your referenced image properly

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\begin{array}{lll}
f(ax+by) &= f(ax) + f(by) &= a f(x) + b f(y)\\
&\uparrow &\uparrow \\
& \textrm{because } f(x+\mu) = f(x) + f(\mu) 
&\textrm{because }f(z\lambda) = z f(\lambda) \\
\end{array}
\]

\end{document}