5.6 Function inverse. Dynamic slides.

Introduction to set theory and to methodology and philosophy of
mathematics and computer programming
Function inverse
An overview
by Jan Plaza
c 2017 Jan Plaza
Use under the Creative Commons Attribution 4.0 International License
Version of November 10, 2017

Note
For a binary relation, the inverse relation is always defined. Recall that the inverse
relation to R is the unique binary relation S, such that for all x, y:
xRy iff ySx
(Such a relation S is denoted R−1.)
For a function f, we want its function inverse g to satisfy
f(x) = y iff g(y) = x
which means that:
g is just the inverse relation to f;
g is a function.
This will require that we start from a “good” function f, not just any function.

Informal Example
We cannot take just any function hoping that its inverse relation will be a function.
Consider function motherOf that maps every person in its domain to their mother.
The inverse relation to motherOf exists and could be called mother-child;
it consists of ordered pairs m, c where m is the mother of c.
mother-child is not a function (because a mother may have two children).
Notice that the function motherOf is not an injection
(as two persons may have the same mother).

Fact
1. If f is a function, and the inverse relation to f is a function,
then f is an injection.
2. If f is an injection,
then the inverse relation to f is a function and an injection.
Deﬁnition
1. Let f be a function. f is invertible or f is an invertible function
if f is an injection.
2. Let f be an invertible function. The function inverse of f or
the inverse function to f , denoted finv , is the inverse relation to f.

Convention
One writes f−1 instead of finv.
Depending on the context, the phrase the inverse of f and the symbol f−1
mean either the relation inverse or function inverse.
The context should suggest which operation is meant.
If in the context functions are discussed, then the function inverse is meant
and there is an implicit claim or assumption that f is invertible.

Example
Conversion from Celsius to Fahrenheit scales of temperature is given by:
f = F(c) =
9
5
c + 32
where c is the temperature in degrees Celsius, and f – in degrees Fahrenheit.
F is an injection from R to R.
How to ﬁnd the inverse function to F?

Example
f = F(c) =
9
5
c + 32
Solve f = 9
5c + 32 for c

Example
f = F(c) =
9
5
c + 32
Solve f = 9
5c + 32 for c: c = . . .

Example
f = F(c) =
9
5
c + 32
Solve f = 9
5c + 32 for c: c = 5
9(f − 32).

Example
f = F(c) =
9
5
c + 32
Solve f = 9
5c + 32 for c: c = 5
9(f − 32). So, what is the inverse function to F?

Example
f = F(c) =
9
5
c + 32
Solve f = 9
5c + 32 for c: c = 5
C(f) =
5
9
(f − 32)

Example
f = F(c) =
9
5
c + 32
Solve f = 9
5c + 32 for c: c = 5
C(f) =
5
9
(f − 32)
Exercise. Complete C−1 = . . .
Exercise. Complete C ◦ F = . . .
Exercise. Complete F ◦ C = . . .

Exercise
Let f(x) = 2x + 1.
Find f−1.
Find (f−1)−1.
Find f ◦ f−1.
Find f−1 ◦ f.

Example
Let f : R −→ R 0 be deﬁned as f(x) = ex.
Let g : R 0 −→ R be deﬁned as g(x) = log x.
g is the inverse function to f.
f is the inverse function to g.
Exercise. Complete g ◦ f = . . .
Exercise. Complete f ◦ g = . . .

Example
Let f : R −→ R 0 be deﬁned as f(x) = x2.
Notice that g : R 0 −→ R deﬁned as g(x) =
√
x, is not an inverse to f;
indeed, (−2)2 = 4, but
√
4=−2.
f does not have an inverse function;
indeed, if h were an inverse function to f,
we would have h(4) = 2 because 22 = 4;
also we would have h(4) = −2 because (−2)2 = 4,
so 2 = h(4) = −2;
this is a contradiction.

Example
Let f : R 0 −→ R 0 be defined as f(x) = x2.
Notice that f is not the same as in the previous example
because the domains are different.
Let g : R 0 −→ R 0 defined as g(x) =
√
x.
g is an inverse function to f;
f is an inverse function to g;
indeed, x2 = y iff x =
√
y, for x, y ∈ R 0.
Exercise. Complete g ◦ f = . . . .
Exercise. Complete f ◦ g = . . . .

Exercise
1. domain(f−1) = . . .

Exercise
1. domain(f−1) = range(f).

Exercise
2. range(f−1) = . . .

Exercise
2. range(f−1) = domain(f).

Exercise
3. True or false? If f : X
1-1
−→ Y , then f−1 : Y
1-1, onto
−→ X.

Exercise
1-1
−→ Y , then f−1 : Y
1-1, onto
−→ X. False.
Find a counter-example.

Exercise
1-1
−→ Y , then f−1 : Y
1-1, onto
−→ X. False.
4. If f : X
1-1
−→ Y , then f−1 :

Exercise
1-1
−→ Y , then f−1 : Y
1-1, onto
−→ X. False.
4. If f : X
1-1
−→ Y , then f−1 : range(f)
1-1, onto
−→ X.

Fact
Let f be a function.
If f is invertible, then f−1 is invertible, and
(f−1)−1 = f.
Corollary
Let f, g be invertible functions. Then,
g = f−1 iﬀ f = g−1.

Proposition
If f is invertible, then f and f−1 are suitable for the following function compositions
and the following equalities hold:
1. f−1 ◦ f = iddomain(f)
2. f ◦ f−1 = idrange(f)
Proposition
Let f : X
1-1
−→ Y and g : Y
1-1
−→ Z.
Then, the following function inverses and function compositions are deﬁned,
and the following equality holds:
(f ◦ g)−1 = g−1 ◦ f−1

Note
Using a tacit assumption that we discuss functions and an earlier convention,
the last few results can stated succinctly, as follows.
1. (f−1)−1 = f
2. f−1 ◦ f = iddomain(f)
3. f ◦ f−1 = idrange(f)
4. Let f : X
1-1
−→ Y and g : Y
1-1
−→ Z; then (f ◦ g)−1 = g−1 ◦ f−1.

5.6 Function inverse. Dynamic slides.

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