Function must be a bijection for its inverse to exist?

The inverse would be ln x.In summary, the concept of inverse functions is defined only for bijections. However, the function y = e^{x} is not bijective, so according to some texts, its inverse (ln x) would not be defined. This is not necessarily true, as it depends on the domain and codomain of the function. In this case, e^{x} and ln x are both bijections when considering the appropriate domain and codomain. Inverse functions can also be defined for injective functions and surjective functions, but they may not be bijective.
  • #1
Bipolarity
776
2
My analysis text defines inverse functions only for bijections.
But [itex] y = e^{x} [/itex] is not bijective, so according to my book it's inverse ([itex] ln x [/itex]) wouldn't be defined? Am I missing something or is my textbook just plain wrong?

I use the text by Bartle and Sherbert.

Thanks!

BiP
 
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  • #2
As I recall it, a function needs only to be injective (one-to-one), but doesn't have to be surjective (onto). Your example of f(x) = ex is one-to-one, and does have an inverse. Likewise, g(x) = ln(x) is one-to-one, and has an inverse as well.

I'd be interested to see why your text might argue that ex and ln(x) don't have inverses.
 
  • #3
I see. That's what I also would expect, but wiki seems to go against :

http://en.wikipedia.org/wiki/Inverse_function

Directly from the above link:

When using codomains, the inverse of a function ƒ: X → Y is required to have domain Y and codomain X. For the inverse to be defined on all of Y, every element of Y must lie in the range of the function ƒ. A function with this property is called onto or a surjection. Thus, a function with a codomain is invertible if and only if it is both injective (one-to-one) and surjective (onto). Such a function is called a one-to-one correspondence or a bijection, and has the property that every element y ∈ Y corresponds to exactly one element x ∈ X.

Perhaps there are many types of definitions for inverses that depend on the context?

BiP
 
  • #4
e^x is a bijection - from the set of real numbers onto the set of positive real numbers. As such it has an inverse function going from the positive reals to the reals - namely, ln(x). Note that ln(x) is also a bijection (from the positive reals onto the reals).

Whenever you talk about a function, you should always keep its domain and codomain in mind. If your function f is defined from X to Y (i.e. you have a function f:X->Y), then an inverse should be a function g:Y->X (note the domain and codomain) such that g(f(x))=f(g(x))=x. This happens if and only if f:X->Y is a bijection.
 
  • #5
surely you have studied functions like arcsin, inverse of sin. since functions must be bijective to have inverses, one must restrict the domain and range until this happens.

i.e. to make sin injective, first we restrict the domain to be [-pi,pi]. Then to make it surjective we restrict the range to be [-1,1].

Then sin has a continuous inverse arcsin, defined on [-1,1] with range in [-pi,pi].

Then if we also want the inverse to be differentiable, we must exclude points where the derivative of sin is zero, so we restrict the domain of sin further to (-pi,pi). Then we have arcsin, a differentiable inverse of sin, defined on (-1,1).Don't they teach this properly in calculus? precalculus? Well to be honest I think our book also did not do so. But that's what professors are for. Maybe the thinking was that freshmen are too young to be frightened by french derived words like injective. But one to one and onto work almost as well.
 
  • #6
It is possible to have only 1-sided inverses, which will be the case if your function is injective but not bijective, and viceversa. If f is 1-1 , then it will have a left inverse g,
i.e., g is such that gof=Id , i.e., gof(x)=x.

Basically, you define g(x) to be the left-inverse, i.e., g(f(x)):=x , for those elements in the range, and define it anyway you want for those elements not in the range.

e.g., take the map from the non-negative integers to themselves, that doubles every number, i.e., f: n-->2n . Then n is 1-1 , but not onto (misses all odd numbers). Define, then g: 2n->n , for even numbers in Z, and any other way for odd numbers. Then gof(n)=n .



A similar argument shows that if f:X-->Y is onto, then f has a right-inverse, i.e., there is h(y) with h:Y-->X , and foh(y)=y. Basically, for all y in Y,the fiber { f^{-1}(y)} of y is non-empty. Map, then , h:y-->yo , i.e., each element of y into an element yo of its fiber*, and , then, almost tautologically, foh(y)=f(yo)=y. Think of , e.g., Sinx from
the reals into [-1,1]. So f(x)=sinx . Define h to send yo in [-1,1] into some element in its fiber :{Sin^{-1}(yo)} .

* Note: if you have not done set theory, feel free to ignore the following:To be more rigorous, you need the axiom of choice to define the right-inverse (not sure about the left inverse; it is too late, i.e., to choose an element in each fiber, when X,Y are infinite). If you're not seeing this, it is not necessary for your understanding.
 
  • #7
you can ALWAYS define the inverse of a function [tex] y=f(x) [/tex]

take the points [tex] (x,f(x)) [/tex] and make a 'reflection' of these points alongside the line

[tex] y=x [/tex] you will get the NUMERICAL inverse of the function.
 
  • #8
zetafunction said:
you can ALWAYS define the inverse of a function [tex] y=f(x) [/tex]

take the points [tex] (x,f(x)) [/tex] and make a 'reflection' of these points alongside the line

[tex] y=x [/tex] you will get the NUMERICAL inverse of the function.

This procedure does not always give you a function though :/
 
  • #9
In order to have an inverse you must have a bijection.

It you think of e[itex]^{x }[/itex]as a map from the real numbers into the real numbers then it has no inverse since negative numbers are not exponentials. But if you think of
e[itex]^{x }[/itex] as a map from the real numbers to the positive numbers then it does have an inverse.
 

Related to Function must be a bijection for its inverse to exist?

1. What is a bijection function?

A bijection function is a type of mathematical function that satisfies two criteria: it is both injective (one-to-one) and surjective (onto). This means that each element in the domain of the function corresponds to exactly one element in the range, and every element in the range has at least one corresponding element in the domain.

2. Why does a function need to be a bijection for its inverse to exist?

If a function is not a bijection, it means that either there are elements in the range that do not have corresponding elements in the domain (not surjective), or there are elements in the domain that correspond to more than one element in the range (not injective). In both cases, the inverse of the function would not be well-defined and would not exist.

3. How do you determine if a function is a bijection?

To determine if a function is a bijection, you can use the horizontal line test. If a horizontal line intersects the graph of the function at more than one point, then the function is not one-to-one and therefore not a bijection.

4. Can a function be a bijection if its domain and range are infinite?

Yes, a function can be a bijection even if its domain and range are infinite. For example, the function f(x) = 2x is a bijection with an infinite domain and range. It satisfies the criteria of being both injective and surjective.

5. Is a bijection function always invertible?

Yes, by definition, a bijection function is always invertible. The inverse function of a bijection is also a bijection. However, it is important to note that not all functions are invertible, and only bijection functions have an inverse that exists.

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