If $F$ is a function and $U$ is a set, then what does $F|U$ mean?
In , the Inverse Function Theorem, the notation is used
Also, the file in stating the Lemma 2.14 states that $A \in U$, what does this mean? A set is an element of another set? What is $A$?
EDIT: Could someone kindly explain to me what is going on in Lemma 2.16. How does $\delta$ come into play? I dont see the role in the lemma. I also don't see why this proves one-to-one
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$\begingroup$It is important to remember a function has three parts: a domain, a range and a rule tying each element of the domain to the range. So if $f: X\rightarrow Y$, and $A\subset X$, $f|A$ is just the function $f: A\rightarrow Y$ so that $(f|A)(x) = f(x)$ for all $x\in A$. When you change the range or domain of a function, you change the function. It's not just the rule!
$\endgroup$ $\begingroup$It means $f$ restricted to the subset $U$. In the paper you look for a local diffeomorphism, so you restrict the function to a suitable subset.
Edit to address later added part of your question "What is $A$": $A$ is a typo. It should read $a$.
$\endgroup$ $\begingroup$If $F: D \rightarrow V$ is a function and $U \subseteq D$, then $F|U$ is the function $f: U \rightarrow V: u \mapsto F(u)$.
$\endgroup$ 2 $\begingroup$$F\mid U$ is the function $F$ restricted to the set $U$, where $U$ is a subset of a larger domain on which $F$ is defined. The image of $F\mid U$ is often denoted by $F[U] = \{f(x): x \in U\} $.
If $A \in U$, then we are talking about $A$ being an element of the subset/set $U$, an element which may or may not be a set itself.
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