TITLE: What does $f|_A$ mean?
QUESTION [8 upvotes]: If $f$ a is a function and $A$ is a set, what could the notation
$$f|_A$$
mean? Is it perhaps "restricted to set $A$"?

REPLY [3 votes]: The notation $f|_A$ is probably best understood via a meaningful example. Before giving one (I hope it will be useful, anyway), it would probably be good to consult two decent references:
1) The Wikipedia page on the restriction of a function.
2) Abstract Algebra by Dummit and Foote (p. 3, 3rd Ed.). 
The relevant portion from the Wiki blurb:

Let $f\colon E\to F$ be a function from a set $E$ to a set $F$, so that the domain of $f$ is in $E$ (i.e., $\operatorname{dom}f\subseteq E$). If $A\subseteq E$, then the restriction of $f$ to $A$ is the function $f|_A\colon A\to F$. 
Informally, the restriction of $f$ to $A$ is the same function as $f$, but is only defined on $A\cap\operatorname{dom} f$.

Wiki's "informal" remark is the key part in my opinion. The following excerpt from Dummit and Foote's Abstract Algebra may be slightly more abstract, but I think a meaningful example will clear everything up.

If $A\subseteq B$, and $f\colon B\to C$, we denote the restriction of $f$ to $A$ by $f|_A$. When the domain we are considering is understood we shall occasionally denote $f|_A$ again simply as $f$ even though these are formally different functions (their domains are different). 
If $A\subseteq B$ and $g\colon A\to C$ and there is a function $f\colon B\to C$ such that $f|_A=g$, we shall say $f$ is an extension of $g$ to $B$ (such a map $f$ need not exist nor be unique). 

Example: Let $g\colon\mathbb{Z}^+\to\{1\}$ be defined by $g(x)=1$ and let $f\colon\mathbb{Z}\setminus\{0\}\to\{1\}$ be defined by $f(x)=\dfrac{|x|}{x}$. Using the notation from the second paragraph above, we have $g\colon A\to C$ and $f\colon B\to C$, where

$A = \mathbb{Z^+}$
$B=\mathbb{Z}\setminus\{0\}$
$C=\{1\}$ 

and, clearly, $A\subseteq B$. Thus, we have the following:
\begin{align}
f|_A &\equiv f\colon\mathbb{Z^+}\to\{1\}\tag{by definition}\\[0.5em]
&= \frac{|x|}{x}\tag{by definition}\\[0.5em]
&= \frac{x}{x}\tag{if $x\in\mathbb{Z^+}$, then $|x|=x$ }\\[0.5em]
&= 1\tag{simplify}\\[0.5em]
&\equiv g\colon\mathbb{Z^+}\to\{1\}\tag{by definition}\\[0.5em]
&= g.
\end{align}
Apart from some slight notational abuse, perhaps, the above example shows that $f$ is an extension of $g$ to $B$ since $f|_A=g$.