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If $F$ is a function and $U$ is a set, then what does $F|U$ mean?

In http://ocw.mit.edu/courses/mathematics/18-101-analysis-ii-fall-2005/lecture-notes/lecture6.pdf, 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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  • $\begingroup$ Where in the file? $\endgroup$ – Asaf Karagila Feb 20 '13 at 1:11
  • $\begingroup$ As a note, this notation is also quite common in linear algebra to denote essentially the same thing: The restriction of a linear operator to an invariant subspace. $\endgroup$ – EuYu Feb 20 '13 at 1:15
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    $\begingroup$ I really really want to interpret your question a different way... $\endgroup$ – John Moeller Feb 20 '13 at 1:19
  • $\begingroup$ Also, for your last question, sets can be elements of other sets. That's not a problem. In fact you can encode most mathematics into sets, so you can always work under the scary assumption that everything is a set. $\endgroup$ – Asaf Karagila Feb 20 '13 at 1:19
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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 $a$.

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If $F: D \rightarrow V$ is a function and $U \subseteq D$, then $F|U$ is the function $f: U \rightarrow V: u \mapsto F(u)$.

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  • $\begingroup$ So all is saying that this $F$ is really just $f: U to V$ map? $\endgroup$ – love Feb 20 '13 at 1:17
  • $\begingroup$ Yes, just $F$ but its domain restricted to $U$. $\endgroup$ – sxd Feb 20 '13 at 2:13
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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!

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$F\mid U$ 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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  • $\begingroup$ You will also sometimes see the notation $f\mid_U$ to denote the restriction of a function $f$ to the subset $U$. $\endgroup$ – Namaste Feb 20 '13 at 1:23
  • $\begingroup$ Also, sometimes there is a little hook on the bar (which I prefer): $f\!\upharpoonright U$ or $f\!\upharpoonright_U$. $\endgroup$ – Nick Matteo May 1 '14 at 21:26

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