EF Inverses

Section 10.5 Inverses

Suppose

f : A \to B

is a function. By definition,

f

associates an element of

B

to each element of

A .

Sometimes we want to reverse this process: given an element

b \in B,

can we determine an element

a \in A

such that

f (a) = b ?

We’ll begin to answer this question by first finding all possible “reverse results” from elements in subsets of

B .

inverse image (of a subset $C$ of the codomain $B$ ): 🔗
the set of all domain elements $a \in A$ for function $f : A \to B$ for which the corresponding output element $f (a)$ lies in the subset $C$ of the codomain
$f^{- 1} (C)$: 🔗

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the inverse image of the subset $C \subseteq B$ under the function $f : A \to B,$ so that

$f^{- 1} (C) = {a \in A | f (a) \in C}$

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Figure 10.5.1. A Venn diagram of a function inverse image.

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Idea 10.5.2.

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As in the figure above,

f^{- 1} (C)

collects together all those elements of

A

whose images under

f

land inside

C .

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Example 10.5.3. Some inverse images under sine.

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Consider

f : R \to R,

f (x) = \sin x .

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Then

f^{- 1} ({- 1, 0, 1}) = {\frac{m π}{2} | m \in Z}

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because

\sin (\frac{m π}{2})

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will equal

0

when

m

is even and will equal

1

- 1

when

m

is odd, and no other input values will produce outputs of

0,

1,

- 1 .

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However,

f^{- 1} ({y \in R | y > 1}) = \emptyset

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because there are no input values for sine that will produce an output value greater than

1 .

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Now let’s return to the question of trying to reverse an input-output relationship

f (a) = b :

the set

f^{- 1} ({b})

collects together all possible candidates for the inverse image of

b .

inverse image (of an element $b$ of the codomain $B$ ): 🔗
the inverse image $f^{- 1} ({b}),$ which consists of all domain elements $a \in A$ for which $f (a) = b$
$f^{- 1} (b)$: 🔗
simplified notation to mean the inverse image of element $b$

🔗

This gives us a way to associate to an element

b \in B

a set

f^{- 1} (b)

of elements of

A .

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Question 10.5.4.

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When does this association

b \mapsto f^{- 1} (b)

give us a function

f^{- 1} : B \to A ?

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There are two possible ways that this will fail to give us a function.

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Suppose there is an element $b \in B$ such that the set $f^{- 1} (b)$ contains (at least) two distinct elements $a_{1}, a_{2} .$ Then in general there is no way to choose between $f^{- 1} (b) = a_{1}$ and $f^{- 1} (b) = a_{2} .$ Therefore, if $f$ is not injective, the function $f^{- 1} : B \to A$ is not well-defined.
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Suppose there is an element $b \in B$ such that $f^{- 1} (b) = \emptyset .$ Then there is no element of $A$ which we can assign to $f^{- 1} (b) .$ Therefore, if $f$ is not surjective, the function $f^{- 1} : B \to A$ is undefined on some elements of $B .$

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So it seems we will need a function to be bijective in order to be able to reverse the input-output rule to obtain an inverse function.

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inverse function: 🔗
for a bijective function $f,$ the inverse function associates to each codomain element of $f$ the corresponding unique domain element that produces it through $f$
$f^{- 1}$: 🔗
the inverse function $f^{- 1} : B \to A$ for bijective function $f : A \to B,$ so that for $b \in B$ we have $f^{- 1} (() b)$ defined to be the unique element $a \in A$ such that $f (a) = b$