5.5 Functions and Sets

Objectives:

• Define the composition of two functions

• Connect our understanding of inverse functions with composition of functions

5.5.1 Connecting Two Major Worlds

Example 5.5.1.

The function $f:ℝ\to ℝ$ given by $f(x)=x^2$ is neither injective nor surjective. Could we restrict the domain and/or codomain of this function to make it bijective?

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On the domain/codomain you chose, what is the function’s inverse function?

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From the above example we see there is great utility in restricting our functions’ domains and codomains. In fact, it is often the only way we can make inverse functions for certain non-bijective functions!

For example, to make non-surjective functions surjective, we merely need to replace the codomain of the function $f$ with $f(X)$. We have used this notation intentionally: this is called the image of the set $X$ under $f$.

Definition 5.5.2.

Let $f:X\to Y$ be a function.

1. If $S\subseteq X$, the image of $S$ under $f$ is defined via

   $$f(S):=\{f(x)\mid x\in S\}.$$

2. If $T\subseteq Y$, the preimage (or inverse image) of $T$ under $f$ is defined via

   $$f^{-1}(T):=\{x\in X\mid f(x)\in T\}.$$

Example 5.5.3.

Let $f(x)=x^2$. Find $f^{-1}([-1,4])$.

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There is an interesting thing we should bring up with this notation: even if a function $f$ is not invertible, we can still take preimages of sets under $f$. The notation $f^{-1}$ can seem a bit unfortunate here, so let us make a quick theorem to help explain the connection between invertibility and preimages.

5.5.2 Proving with Functions and Sets

Theorem 5.5.4.

Let $f:X\to Y$ be a bijective function. Then for all values $y\in Y$,

$$f^{-1}(\{y\})=\{f^{-1}(y)\}.$$

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The notation on the left-hand side denotes the preimage of the set $\{y\}$ under $f$; the notation on the right-hand side denotes the output of $y$ under the inverse function $f^{-1}$. Notice that pre-images always are sets, not numbers, even if there is one element in this set.

Example 5.5.5.

1. Show that the following statement is true: for any function $f:X\to Y$ where $C,D\subset Y$, $f^{-1}(C)\cap f^{-1}(D)\subseteq f^{-1}(C\cap D)$.

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2. Define $f:ℝ\to ℝ$ via $f(x)=x^2$. Find two nonempty subsets $A$ and $B$ of $ℝ$ such that $A\cap B=\mathrm{\varnothing }$ but $f(A)=f(B)$.

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   Using this, show that the following statement is false: for any function $f:X\to Y$ where $A,B\subset X$, $f(A)\cap f(B)\subseteq f(A\cap B)$.

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Theorem 5.5.6.

Let $f:X\to Y$ be an injective function and let $S\subset X$. Then $f^{-1}(f(S))=S$.

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Theorem 5.5.7.

Let $g:X\to Y$ be a surjective function and let $T\subset Y$. Then $f(f^{-1}(T))=T$.

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Theorem 5.5.8.

Let $f:X\to Y$ be a function and suppose that ${(A_{\alpha })}_{\alpha \in \mathrm{\Delta }}$ is a collection of subsets of $Y$. Then we have the following:

1. it is true that $f\left({\bigcap }_{\alpha \in \mathrm{\Delta }}{A}_{\alpha }\right)\subseteq {\bigcap }_{\alpha \in \mathrm{\Delta }}f(A_{\alpha })$.

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2. it is not necessarily true that ${\bigcap }_{\alpha \in \mathrm{\Delta }}f(A_{\alpha })\subseteq f\left({\bigcap }_{\alpha \in \mathrm{\Delta }}{A}_{\alpha }\right)$.

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