# Images and Inverses

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.

• The first definition doesn't look quite right. – Lucas Henrique Jan 9 '19 at 5:23
• You're right! I made a typo error. I've updated it. – Ryan Jan 9 '19 at 5:26
• You may define $f(T)=\{f(t):t\in T\}$ – Shubham Johri Jan 9 '19 at 6:55
• @ShubhamJohri this is not good, especially at the start, because in first glance there is no reason for this to be a set – ℋolo Jan 9 '19 at 8:45
• Beg your pardon? – Shubham Johri Jan 9 '19 at 9:36

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 $$f.$$

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.

• I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me. – Lucas Henrique Jan 9 '19 at 5:32
• 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. – Ryan Jan 9 '19 at 5:33
• $f(\Bbb R)=\Bbb R^+\cup\{0\}$ – Shubham Johri Jan 9 '19 at 6:15

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.

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.