Why does an inverse exist only for surjective functions?

In summary, a function must be 1-1 and onto (or surjective) in order to have an inverse. The inverse can only be defined on the range of the function, and if the range is not equal to the codomain, then the inverse may not exist. This can be seen in the example of the exponential function, which is 1-1 but its inverse is only defined on a subset of the codomain. Therefore, if a function's range is not equal to its codomain, its inverse may not exist within the set of functions that map from the domain to the codomain.
  • #1
kay
60
2
In other words, why does a function have to be onto (or surjective, i.e. Range=Codomain) for its inverse to exist?
 
Mathematics news on Phys.org
  • #2
kay said:
In other words, why does a function have to be onto (or surjective, i.e. Range=Codomain) for its inverse to exist?

The key property for a function to have an inverse is that it is 1-1. Any 1-1 function will have an inverse. The inverse, however, can only be defined on the range of the function. Perhaps this is best explained by an example:

The exponential function ##e^x## maps the real number line ##(-\infty, \infty)## onto ##(0, \infty)##. The exponential is 1-1, so there exists an inverse (the natural logarithm). But, the inverse is only defined on ##(0, \infty)##. You can't define the inverse function on all of the real number line.

This raises perhaps a technical point that if you are considering the set of functions that map ##\mathbb{R}## to ##\mathbb{R}## then the exponential function is in this set, but there is no inverse within this set of functions. From that point of view, the inverse doesn't exist.
 
  • Like
Likes kay
  • #3
PeroK said:
The key property for a function to have an inverse is that it is 1-1. Any 1-1 function will have an inverse. The inverse, however, can only be defined on the range of the function. Perhaps this is best explained by an example:

The exponential function ##e^x## maps the real number line ##(-\infty, \infty)## onto ##(0, \infty)##. The exponential is 1-1, so there exists an inverse (the natural logarithm). But, the inverse is only defined on ##(0, \infty)##. You can't define the inverse function on all of the real number line.

This raises perhaps a technical point that if you are considering the set of functions that map ##\mathbb{R}## to ##\mathbb{R}## then the exponential function is in this set, but there is no inverse within this set of functions. From that point of view, the inverse doesn't exist.
Actually my question was, that why can't an inverse exist for a function whose range is not necessarily equal to but is rather a subset of the codomain.
 
  • #4
kay said:
Actually my question was, that why can't an inverse exist for a function whose range is not necessarily equal to but is rather a subset of the codomain.

An inverse of a function [itex]f: A \to B[/itex] is a function [itex]g: B \to A[/itex] such that [itex]g(f(a)) = a[/itex] for all [itex]a \in A[/itex] and [itex]f(g(b)) = b[/itex] for all [itex]b \in B[/itex]. An inverse is unique if it exists.

Let [itex]f: \{1,2\} \to \{1,2,3\} : x \mapsto x[/itex]. Now the inverse of [itex]f[/itex], if it exists, is a function [itex]g: \{1,2,3\} \to \{1,2\}[/itex]. Now we must have [itex]g(1) = 1[/itex] and [itex]g(2) = 2[/itex], but what value do you assign to [itex]g(3)[/itex] such that [itex]f(g(3)) = 3[/itex]?
 
  • Like
Likes kay
  • #5
pasmith said:
An inverse of a function [itex]f: A \to B[/itex] is a function [itex]g: B \to A[/itex] such that [itex]g(f(a)) = a[/itex] for all [itex]a \in A[/itex] and [itex]f(g(b)) = b[/itex] for all [itex]b \in B[/itex]. An inverse is unique if it exists.

Let [itex]f: \{1,2\} \to \{1,2,3\} : x \mapsto x[/itex]. Now the inverse of [itex]f[/itex], if it exists, is a function [itex]g: \{1,2,3\} \to \{1,2\}[/itex]. Now we must have [itex]g(1) = 1[/itex] and [itex]g(2) = 2[/itex], but what value do you assign to [itex]g(3)[/itex] such that [itex]f(g(3)) = 3[/itex]?
I got it. Thanks a lot for your help. :)
 

Related to Why does an inverse exist only for surjective functions?

1. Why does an inverse exist only for surjective functions?

An inverse function is a function that "undoes" the action of another function. In order for an inverse function to exist, each element in the range of the original function must have a unique preimage. This means that every output of the function must have a corresponding input. Surjective functions, also known as onto functions, have this property, whereas injective and bijective functions do not.

2. What is the difference between a surjective function and an injective function?

A surjective function is a function where every element in the range has a unique preimage, or input. This means that the function covers its entire range. An injective function, on the other hand, is a function where each element in the domain has a unique output. In other words, the function does not map two different inputs to the same output.

3. Can a function be both surjective and injective?

Yes, a function can be both surjective and injective, in which case it is known as a bijective function. This type of function has a unique input for every output and vice versa. In other words, it is a one-to-one correspondence between the domain and the range.

4. How can I determine if a function is surjective?

To determine if a function is surjective, you can check if every element in the range has a corresponding input in the domain. One way to do this is to graph the function and see if the entire range is covered. Another way is to see if you can solve for the input given any output in the range. If you can, then the function is surjective.

5. Are all surjective functions invertible?

No, not all surjective functions are invertible. In order for a function to have an inverse, it must be both surjective and injective, meaning it must cover its entire range and have a unique input for every output. If a function is only surjective, it may have multiple inputs for a single output, making it non-invertible.

Similar threads

Replies
1
Views
1K
  • Topology and Analysis
Replies
8
Views
1K
Replies
17
Views
2K
Replies
3
Views
325
Replies
16
Views
1K
Replies
11
Views
589
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
Replies
18
Views
5K
Replies
11
Views
1K
Back
Top