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Function must be a bijection for its inverse to exist?

  1. Mar 12, 2012 #1
    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.


  2. jcsd
  3. Mar 12, 2012 #2


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    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.
  4. Mar 12, 2012 #3
    I see. That's what I also would expect, but wiki seems to go against :


    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?

  5. Mar 12, 2012 #4


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    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.
  6. Mar 12, 2012 #5


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    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.
  7. Mar 13, 2012 #6


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    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.
  8. Mar 15, 2012 #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.
  9. Mar 15, 2012 #8


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    This procedure does not always give you a function though :/
  10. Mar 16, 2012 #9


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    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.
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