6.9: Image and Pre-Image

It is sometimes necessary to collect together many values of a function. Let us start with a real-world example.

Example \(6.9.1\).

Suppose the astronomy club at an elementary school decides to have a Father’s Day party. Then they should make a list of all of their fathers, so that invitations can be sent. Mathematically speaking, they want to make a set that contains precisely the people who are the father of someone in the club. That is, if A1 is the set of people in the club, then they are interested in \[\left\{x \mid \exists a \in A_{1},(x=\text { father }(a))\right\} .\]

Another way of thinking of this is that they should apply the \(\text {father}\) function to every element of the set \(A_{1}\), and gather all of the resulting values into a set. The mathematical notation for this set that gathers together the values is \[\left\{\text { father }(a) \mid a \in A_{1}\right\} .\]

In English, we could call this set “the fathers of the elements of \(A_{1}\),” but mathematicians abbreviate this to “\(\text {father} (A_{1})\).” In summary, the same set has three names: \[\text { father }\left(A_{1}\right)=\left\{\text { father }(a) \mid a \in A_{1}\right\}=\left\{x \mid \exists a \in A_{1},(x=\text { father }(a))\right\} .\]

A similar idea can be applied to any function \(f : A \rightarrow B\). Namely, if \(A_{1} \subset A\), then we can apply \(f\) to every element of the set \(A_{1}\), and gather all of the resulting values into a set. We call this set \(f(A_{1})\).

Definition \(6.9.2\).

Suppose \(f : A \rightarrow B\), and \(A_{1} \subset A\). The image of \(A_{1}\) under \(f\) is \[f\left(A_{1}\right)=\left\{f(a) \mid a \in A_{1}\right\} .\]

It is a subset of \(B\). The notation means that, for all \(x\), we have \[x \in f\left(A_{1}\right) \quad \Leftrightarrow \quad \exists a \in A_{1},(x=f(a)).\]

Remark \(6.9.3\).

We can take the image of any subset of the domain of \(f\), and the result will be some subset of the range of \(f\). In the special case where we take the entire domain of \(f\) as our set \(A_{1}\), we obtain the entire range of \(f\) as the image.

You are expected to be able to combine the definition of “image” with the proof techniques that you already know.

Example \(6.9.4\).

Assume \(f : A \rightarrow B\). Show that if \(A_{1}\) and \(A_{2}\) are subsets of \(A\), and \(f\) is one-to-one, then \[f\left(A_{1}\right) \cap f\left(A_{2}\right) \subset f\left(A_{1} \cap A_{2}\right) .\]

Solution

Given \(b \in f\left(A_{1}\right) \cap f\left(A_{2}\right)\), we know \(b \in f\left(A_{1}\right)\) and \(b \in f\left(A_{2}\right)\). Therefore, since \(b \in f\left(A_{1}\right)\), we know there is some \(a_{1} \in f\left(A_{1}\right)\), such that \(b = f(a_{1})\). Also, since \(b \in f\left(A_{2}\right)\), we know there is some \(a_{2} \in f\left(A_{2}\right)\), such that \(b = f(a_{2})\). Then \[f\left(a_{1}\right)=b=f\left(a_{2}\right) .\]

Since \(f\) is one-to-one, this implies \(a_{1} = a_{2} \in A_{2}\). Since we also know that \(a_{1} \in A_{1}\), this implies \(a_{1} \in A_{1} \cap A_{2}\). So \(f(a_{1}) \in f(A_{1} \cap A_{2})\). Since \(b = f(a_{1})\), this means \(b \in f(A_{1} \cap A_{2})\). Since \(b\) is an arbitrary element of \(f(A_{1}) \cap f(A_{2})\), we conclude that \[f\left(A_{1}\right) \cap f\left(A_{2}\right) \subset f\left(A_{1} \cap A_{2}\right) .\]

Exercise \(6.9.5\).

Assume \(f : A \rightarrow B\).

1. Show that if \(A_{1}\) and \(A_{2}\) are subsets of \(A\), such that \(A_{2} \subset A_{1}\), then \(f(A_{2}) \subset f(A_{1})\).
2. Assume \(f\) is one-to-one, and \(a \in A\). Show that if \(f(a) \in f(A_{1})\), then \(a \in A_{1}\).

Taking the image of a subset of the domain yields a subset of the codomain. Sometimes we need to go the other direction.

Example \(6.9.6\).

Perhaps we would like to make a list of all the people whose father is a friend of the pop singer Bono. If \(B_{1}\) is the set of Bono’s friends, then the mathematical notation for the set of these people is \[\left\{x \in \text { PEOPLE } \mid \text { father }(x) \in B_{1}\right\} .\]

Notice that if the \(\text {father}\) function had an inverse, then the same set could be obtained by applying \(\text{father}^{−1}\) to the elements of \(B_{1}\). That is, the set would be \(\text{father}^{−1}(B_{1})\). Mathematicians use this notation for the set even if there is no inverse function.

Definition \(6.9.7\).

Suppose \(f : A \rightarrow B\), and \(B_{1} \subset B\). The pre-image (or inverse image) of \(B_{1}\) under \(f\) is \[f^{-1}\left(B_{1}\right)=\left\{a \in A \mid f(a) \in B_{1}\right\} .\]

It is a subset of \(A\). When \(B_{1} = \{b\}\) has only one element, we usually write \(f^{−1}(b)\), instead of \(f^{−1} (\{b\})\).

WARNING. The fact that we write \(f^{−1} (B_{1})\) does not imply that \(f\) has an inverse, or that \(f^{−1}\) is a function. This is simply a notation that refers to the set we have defined.

Example \(6.9.8\).

1. For the function mother: \(\text {PEOPLE } \rightarrow \text { WOMEN}\), \(\text {mother^{−1} (m)\) is the set of all children of \(m\).
2. For the function \(f : \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x) = x^{2}\):
   1. We have \(f^{−1}(4) = \{2, −2\}\), because 2 and −2 are all of the square roots of 4.
   2. We have \(f^{−1}([0, 4]) = [−2, 2]\), because \(0 \leq x^{2} \leq 4 \text { iff } −2 \leq x \leq 2\).

Here are examples of proofs involving inverse images:

Example \(6.9.9\).

Suppose \(f: A \rightarrow B\) and \(B_{1} \subset B\).

1. We have \(f(f^{−1}(B_{1})) \subset B_{1}\).
2. If \(f\) is onto, then \(f(f^{−1}(B_{1})) = B_{1}\).

Solution

1. Let \(b \in f\left(f^{-1}\left(B_{1}\right)\right)\). By definition, we have \[f\left(f^{-1}\left(B_{1}\right)\right)=\left\{f(a) \mid a \in f^{-1}\left(B_{1}\right)\right\} ,\]
   so we must have \(b = f(a_{1})\), for some \(a_{1} \in f^{−1}(B_{1})\). From the definition of \(f^{−1}(B_{1})\), we know that \(f(a_{1}) \in B_{1}\). Therefore \(b = f(a_{1}) \in B_{1}\). Since \(b\) is an arbitrary element of \(f(f^{−1}(B_{1}))\), this implies that \(f(f^{−1}(B_{1})) \subset B_{1}\), as desired.
2. Assume \(f\) is onto. We know, from (1), that \(f(f^{−1}(B_{1})) \subset B_{1}\), so it suffices to show that \(B_{1} \subset f\left(f^{-1}\left(B_{1}\right)\right)\).

Let \(b \in B_{1}\) be arbitrary. Because \(f\) is onto, we know there exists \(a_{1} \in A\), such that \(f(a_{1}) = b\). Then \(f(a_{1}) = b \in B_{1}\), so \(a_{1} \in f^{−1}(B_{1})\). Therefore \[f\left(a_{1}\right) \in\left\{f(a) \mid a \in f^{-1}\left(B_{1}\right)\right\}=f\left(f^{-1}\left(B_{1}\right)\right) .\]

Since \(f(a_{1}) = b\), we conclude that \(b \in f\left(f^{-1}\left(B_{1}\right)\right)\). Since \(b\) is an arbitrary element of \(B_{1}\), this implies that \(B_{1} \subset f\left(f^{-1}\left(B_{1}\right)\right)\), as desired.

Exercise \(6.9.10\).

Suppose that \(f: A \rightarrow B\), that \(A_{1} \subset A\), and that \(B_{1} \subset B\).

1. Show that if \(B_{2} \subset B_{1}\), then \(f^{-1}\left(B_{2}\right) \subset f^{-1}\left(B_{1}\right)\).
2. Show \(A_{1} \subset f^{-1}\left(f\left(A_{1}\right)\right)\).

Exercise \(6.9.11\).

Assume \(f: X \rightarrow Y\), \(A \subset Y\), and \(B \subset Y\). Show \[f^{-1}(A) \cap f^{-1}(B)=f^{-1}(A \cap B) .\]

Exercise \(6.9.12\).

Assume \(f: X \rightarrow Y\), \(g: Y \rightarrow Z\), \(X_{1} \subset X\), \(Z_{1} \subset Z\), and \((g \circ f)\left(X_{1}\right) \subset Z_{1}\). Show \(f(X_{1}) \subset g^{−1}(Z_{1})\).