5.6 Function inverse. A handout.
- 1. 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
- 2. 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.
- 3. 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).
- 4. 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.
- 5. 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.
- 6. 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? Solve f = 9 5c + 32 for c: c = . . . 5 9(f − 32). So, what is the inverse function to F? C(f) = 5 9 (f − 32) Exercise. Complete C−1 = . . . Exercise. Complete C ◦ F = . . . Exercise. Complete F ◦ C = . . .
- 7. Exercise Let f(x) = 2x + 1. Find f−1. Find (f−1)−1. Find f ◦ f−1. Find f−1 ◦ f.
- 8. 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 = . . .
- 9. 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.
- 10. 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 = . . . .
- 11. Exercise 1. domain(f−1) = . . . range(f). 2. range(f−1) = . . . domain(f). 3. True or false? If f : X 1-1 −→ Y , then f−1 : Y 1-1, onto −→ X. False. Find a counter-example. 4. If f : X 1-1 −→ Y , then f−1 : . . . range(f) 1-1, onto −→ X.
- 12. 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.
- 13. 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
- 14. 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.