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Definition 0.2.10 Let $A$ and $B$ be sets, and $f: A \rightarrow B$ be a function. Let $S \subseteq B$. Then the set $f^{-1}(S)=\{ x\in A : f(x) \in S\}$ is called the inverse image of the set $S$ under the function $f$.

Definition 0.2.14 Let $A$ and $B$ be sets, and $f: A \rightarrow B$ be a function. Let $T \subseteq A$. Then the set $f(T)=\{x\in B : x=f(t) \text{ for some }t \in T \}$ is called the image of the set $T$ under the function $f$.

How do these two definitions differ from traditional usage of $f$ and $f^{-1}$ as in a calculus course?

My attempt to make any sense of it:

They are similar in many ways. In the definitions, for a function $f$ that maps $A$ to $B$ is similar to plugging a value $x$ into $f(x)$ and getting an output $y$. But, in the definitions given above, $f$ is invertible if it is one-to-one and onto, but how does this differ from the traditional $f$ used in calculus? I cannot seem to find the ties or correlations.

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    $\begingroup$ The first definition doesn't look quite right. $\endgroup$ – Lucas Henrique Jan 9 at 5:23
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    $\begingroup$ You're right! I made a typo error. I've updated it. $\endgroup$ – Ryan Jan 9 at 5:26
  • $\begingroup$ You may define $f(T)=\{f(t):t\in T\}$ $\endgroup$ – Shubham Johri Jan 9 at 6:55
  • $\begingroup$ @ShubhamJohri this is not good, especially at the start, because in first glance there is no reason for this to be a set $\endgroup$ – Holo Jan 9 at 8:45
  • $\begingroup$ Beg your pardon? $\endgroup$ – Shubham Johri Jan 9 at 9:36
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It is common and convenient in many subjects to assume that some set $A$ that we are interested in is an anti-transitive set: That no member of $A$ is a subset of $A.\, $ E.g. we usually assume that a subset of $\Bbb R$ is never a member of $\Bbb R,$ so that if $f:\Bbb R \to \Bbb R$ and $T\subset \Bbb R$ then $f(T)=\{f(x):x\in T\}$ and $f^{-1}(T)=f^{-1}T=\{x:f(x)\in T\}$ are unambiguous definitions. And if $A,B$ are anti-transitive sets and $f:A\to B$ is one-to-one (injective) and if $b$ belongs to the (unambiguous) set $f(A)$, then defining $a=f^{-1}(b)\iff f(a)=b$ is also unambiguous. The anti-transitive assumption occurs, often tacitly, in many books and papers.

Difficulties arise, especially in Set Theory, when we cannot safely assume anti-transitivity. E.g. if $\emptyset$ and $\{\emptyset\}$ both belong to the domain of a function $f$, then it is unclear what $f(\{\emptyset \})$ "should" mean. In Set Theory a function $f:A\to B$ $is$ its graph, so $f$ is a certain type of subset of $A\times B.$ And in Set Theory it is preferable to write $b=f(a)$ to only mean that $(a,b)\in f.$ And then the notation $f''T$ (read $f$-double-prime-$T$) is used to denote $\{f(t): t\in T\cap dom(f)\},$ the image of $T$ under $f.$

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Usually when we think about the image of a function, we imagine the whole function. What the author defined here is a generalization: let $f(x) = x^2$, for example; then $f[\Bbb R] = \Bbb R_{\geq 0}$ as expected, but also we can look specifically at some part of the graph, like e.g. $f[[-1,1]] = [0;1]$. Try to sketch those in your mind: it's like "cropping" the function to those specific interesting parts to see what they look like. The same for the inverse image: you're looking for all the values that make your function output something in your set.

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  • $\begingroup$ I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me. $\endgroup$ – Lucas Henrique Jan 9 at 5:32
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    $\begingroup$ Thank you for the response. Your response is fine, as I am really trying to digest most of the material in this chapter. Thanks again. $\endgroup$ – Ryan Jan 9 at 5:33
  • $\begingroup$ $f(\Bbb R)=\Bbb R^+\cup\{0\}$ $\endgroup$ – Shubham Johri Jan 9 at 6:15
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The inverse image is a function from the power set of the codomain to the power set of the domain. That is,$$f^{-1}:P(B)\to P(A)$$that operates on a subset of the codomain, say $S$, to return a subset of the domain, say $M$, containing values that map to some member of $S$ under $f$. It is not required for each member of $S$ to have a pre-image. At the same time, some members of $S$ can have more than one pre-image, in case $f$ is not injective.

That is to say, $f$ needn't be invertible for defining the inverse image of a subset of its codomain. For example, consider the function $f:\{1,2,3\}\to\{1,2,3\},f=\{(1,1),(2,1),(3,2)\}$. Clearly, $f$ is not invertible since it is neither injective nor surjective. But,

  • $f^{-1}(\phi)=f^{-1}(\{3\})=\phi$, since no element maps to $3$.

  • $f^{-1}(\{1\})=\{1,2\}$, since both $1,2$ map to $1$.

  • $f^{-1}(\{1,3\})=\{1,2\}$, since both $1,2$ map to $1$ while no element maps to $3$.

  • $f^{-1}(\{2\})=\{3\}$, since only $3$ maps to $2$.

The direct image, or just the image, is a function from the power set of the domain to the power set of the range, or codomain, of $f$. That is,$$f:P(A)\to P(B)$$is a function that takes in a subset of the domain, say $T$, and returns the set $R$ containing the images of all the elements of $T$. In the example given above, we have

  • $f(\phi)=\phi$

  • $f(\{1\})=f(\{2\})=\{1\}$, since both $1,2$ map to $1$.

  • $f(\{3\})=\{2\}$, since $3$ maps to $2$.

  • $f(\{1,3\})=\{1,2\}$, since $1$ maps to $1$ and $3$ maps to $2$.

Remark. Although both the function and the direct image use the notation $f$, you can detect if it is the latter that is being talked about by the argument of $f$. In case of direct image, the argument will be a set, such as $\{1,2\}$. Similarly for the inverse of the function, if it exists, and the inverse image.

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why will there be any difference it seems the $f$ and $ f^{-1} $ defined here is exactly as that of calculus . one thing you may say as a difference is in calculus most generally we restrict ourselves to metric spaces as the two sets.

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