The notion of an inverse function arises from the natural question: given a function that assigns to each element of its domain an element of the codomain, can this correspondence be followed in the opposite direction?
In other words, if a function \(f:A\to B\) assigns to an element \(x\in A\) the value \(y=f(x)\in B\), one may ask whether, knowing \(y\), it is possible to recover unambiguously the element \(x\) it came from.
This is not always possible. A function may take the same value at distinct points of its domain, or it may fail to reach every element of the codomain. For this reason the inverse function depends not only on the formula defining the function, but also on the domain, the codomain, and the properties of injectivity and surjectivity.
Our aim is to make clear when a function admits an inverse, how the inverse function is defined rigorously, and what role the notions of right inverse and left inverse play. The latter let us understand more precisely what happens when a function is not bijective but retains only one of the two fundamental properties: either injectivity or surjectivity.
Contents
- What it means to invert a function
- Definition of the inverse function
- When the inverse function exists
- Uniqueness of the inverse and the role of domain and codomain
- How to find the inverse function
- Examples of invertible and non-invertible functions
- Left inverse of a function
- Right inverse of a function
- The relationship between inverse, left inverse, and right inverse
- Summary
What it means to invert a function
Let
\[ f:A\to B \]
be a function. By definition, to each element \(x\in A\) the function assigns one and only one element \(f(x)\in B\).
To invert a function means to ask whether this assignment can be traversed the other way around: no longer starting from \(x\) to obtain \(f(x)\), but starting from a value \(y\in B\) and recovering the element \(x\in A\) that produced it.
In symbols, if
\[ y=f(x), \]
we ask whether \(x\) can be determined from \(y\).
This operation, however, is not always possible. The first obstacle arises when two distinct elements of the domain share the same image. If there exist \(x_1,x_2\in A\), with \(x_1\ne x_2\), such that
\[ f(x_1)=f(x_2), \]
then, starting from the value \(f(x_1)=f(x_2)\), we cannot tell unambiguously whether the original element was \(x_1\) or \(x_2\). In this case the inversion cannot give rise to a function, since a function must assign to each element of its domain one and only one value.
A second obstacle arises when some elements of the codomain are not values taken by the function. If there exists an element \(y\in B\) that does not belong to the image of \(f\), then there is no \(x\in A\) such that
\[ f(x)=y. \]
In this case, starting from \(y\), we cannot recover any element of the domain.
Inverting a function therefore requires two fundamental conditions: every element of the codomain must actually be attained by the function, and it must be attained by exactly one element of the domain. The first requirement is surjectivity, the second is injectivity.
When both conditions hold, the function establishes a one-to-one correspondence between the elements of \(A\) and the elements of \(B\). Only in this situation can a genuine inverse function be defined, one that assigns to each element of \(B\) the unique element of \(A\) from which it comes.
Definition of the inverse function
Let
\[ f:A\to B \]
be a function. A function
\[ f^{-1}:B\to A \]
is called the inverse of \(f\) if, for every \(x\in A\) and every \(y\in B\), the following equivalence holds:
\[ y=f(x) \quad \Longleftrightarrow \quad x=f^{-1}(y). \]
In other words, the inverse function assigns to each element \(y\in B\) the unique element \(x\in A\) whose image under \(f\) is precisely \(y\).
When it exists, the inverse function reverses the direction of the correspondence defined by \(f\). If the original function sends \(x\) to \(y\), then the inverse function sends \(y\) to \(x\):
\[ f(x)=y \quad \Longrightarrow \quad f^{-1}(y)=x. \]
This notation must be read with care. The symbol \(f^{-1}\) does not denote the reciprocal of the function \(f\). In general,
\[ f^{-1}(x)\ne \frac{1}{f(x)}. \]
Rather, the symbol \(f^{-1}\) denotes the function that performs the operation inverse to \(f\), that is, the function carrying each value of the codomain back to the element of the domain from which it came.
The characteristic property of the inverse function can also be expressed through composition of functions. If \(f:A\to B\) admits an inverse \(f^{-1}:B\to A\), then the identities
\[ f^{-1}\circ f=\operatorname{id}_A \]
and
\[ f\circ f^{-1}=\operatorname{id}_B \]
hold.
The first identity means that, starting from an element of \(A\) and applying first \(f\) and then \(f^{-1}\), we return to the original element:
\[ f^{-1}(f(x))=x \qquad \text{for every } x\in A. \]
The second identity means that, starting from an element of \(B\) and applying first \(f^{-1}\) and then \(f\), we return to the original element:
\[ f(f^{-1}(y))=y \qquad \text{for every } y\in B. \]
These two equalities capture rigorously the meaning of the inverse function: applying a function and then its inverse, or applying the inverse first and then the function, leaves the original element unchanged.
When the inverse function exists
Not every function admits an inverse. For a function
\[ f:A\to B \]
to admit an inverse
\[ f^{-1}:B\to A, \]
every element of \(B\) must come from one and only one element of \(A\).
More precisely, for every \(y\in B\) there must exist a unique \(x\in A\) such that
\[ f(x)=y. \]
This condition involves two separate requirements.
The first is an existence requirement: for every \(y\in B\) there must exist at least one element \(x\in A\) with \(f(x)=y\). In this way every element of the codomain is actually attained by the function. This is precisely surjectivity.
The second is a uniqueness requirement: for every \(y\in B\) there must exist at most one element \(x\in A\) with \(f(x)=y\). In this way no two distinct elements of the domain can have the same image. This is precisely injectivity.
Consequently, a function admits an inverse if and only if it is both injective and surjective, that is, if and only if it is bijective.
In symbols:
\[ f:A\to B \text{ admits an inverse } f^{-1}:B\to A \quad \Longleftrightarrow \quad f \text{ is bijective}. \]
The necessity of this condition can be seen directly. If \(f\) is not injective, there exist two distinct elements \(x_1,x_2\in A\) such that
\[ f(x_1)=f(x_2). \]
In that case, starting from the common value \(f(x_1)=f(x_2)\), any putative inverse would have to assign to the same element of \(B\) both \(x_1\) and \(x_2\). This is impossible, since a function must assign a single value to each element of its domain.
If, on the other hand, \(f\) is not surjective, there is at least one element \(y\in B\) that is the image of no element of \(A\). For such a \(y\), there is no \(x\in A\) such that
\[ f(x)=y. \]
In this case the inverse could not be defined on all of \(B\), since no element of \(A\) would correspond to that element \(y\).
Conversely, if \(f\) is bijective, then for every \(y\in B\) there exists one and only one \(x\in A\) such that \(f(x)=y\). We may therefore define
\[ f^{-1}(y)=x. \]
This definition is well posed: the existence of \(x\) is guaranteed by surjectivity, while its uniqueness is guaranteed by injectivity.
Hence bijectivity is not merely a sufficient condition for the existence of the inverse function: it is also a necessary one.
Uniqueness of the inverse and the role of domain and codomain
An invertible function establishes a one-to-one correspondence between the elements of the domain and those of the codomain.
If
\[ f:A\to B \]
is bijective, then every element \(y\in B\) is the image of one and only one element \(x\in A\). Consequently, the inverse
\[ f^{-1}:B\to A \]
is the function that assigns to each \(y\in B\) that unique element \(x\in A\) such that
\[ f(x)=y. \]
In symbols:
\[ f^{-1}(y)=x \quad \Longleftrightarrow \quad f(x)=y. \]
The inverse function is therefore not an object artificially attached to the original function: it is uniquely determined by \(f\), whenever \(f\) is bijective.
Indeed, if there were two inverses \(g:B\to A\) and \(h:B\to A\) of the same function \(f\), then for every \(y\in B\) we would have
\[ f(g(y))=y \qquad \text{and} \qquad f(h(y))=y. \]
Since \(f\) is injective, the equality
\[ f(g(y))=f(h(y)) \]
forces
\[ g(y)=h(y). \]
This holds for every \(y\in B\), so \(g=h\). The inverse of a function, when it exists, is therefore unique.
Moreover, if \(f:A\to B\) is bijective with inverse \(f^{-1}:B\to A\), then \(f^{-1}\) is itself bijective, and its inverse is the original function:
\[ (f^{-1})^{-1}=f. \]
This equality expresses the fact that inverting the correspondence a second time brings us back to the original function.
Invertibility always depends on the function taken together with its domain and codomain. The same formula may define an invertible or a non-invertible function, depending on the sets between which it is considered: one must therefore always specify the domain, the codomain, and the rule of assignment.
For example, the function
\[ f:\mathbb R\to \mathbb R,\qquad f(x)=x^2 \]
is not invertible, because it is not injective: indeed \(f(-1)=f(1)=1\).
If instead we consider the function
\[ f:[0,+\infty)\to [0,+\infty),\qquad f(x)=x^2, \]
then \(f\) is bijective and therefore admits an inverse. In this case
\[ f^{-1}(x)=\sqrt{x}. \]
This example shows that the formula alone is not enough: to decide whether a function is invertible we must take into account its domain, its codomain, and its properties.
How to find the inverse function
When a function is invertible, its inverse can often be found starting from the equation that defines the function.
Suppose we have a function
\[ f:A\to B \]
given by some expression \(y=f(x)\). To determine the inverse, we solve the equation
\[ y=f(x) \]
for the variable \(x\).
If the function is invertible, then for every \(y\in B\) there is one and only one \(x\in A\) with \(f(x)=y\). Hence solving the equation for \(x\) yields an expression of the form
\[ x=f^{-1}(y). \]
At this point, if we wish to write the inverse using \(x\) as the independent variable, we may rename the variable \(y\) as \(x\). This step is merely a relabeling of the variable, not a change in mathematical meaning.
In practice, the procedure is as follows:
- write \(y=f(x)\);
- solve the equation for \(x\);
- obtain \(x=f^{-1}(y)\);
- relabel the independent variable, writing \(f^{-1}(x)\) in place of \(f^{-1}(y)\).
Consider, for example, the function
\[ f:\mathbb R\to \mathbb R,\qquad f(x)=2x+3. \]
The function is bijective, hence it admits an inverse. We write
\[ y=2x+3. \]
Solving for \(x\):
\[ y-3=2x, \]
so that
\[ x=\frac{y-3}{2}. \]
Therefore
\[ f^{-1}(y)=\frac{y-3}{2}. \]
Relabeling the independent variable, we obtain
\[ f^{-1}(x)=\frac{x-3}{2}. \]
It is always advisable to check the result by composition. Indeed:
\[ f^{-1}(f(x))=f^{-1}(2x+3)=\frac{2x+3-3}{2}=x \]
for every \(x\in\mathbb R\), and
\[ f(f^{-1}(x))=f\left(\frac{x-3}{2}\right)=2\cdot\frac{x-3}{2}+3=x \]
for every \(x\in\mathbb R\).
The two identities confirm that the function we found is indeed the inverse of \(f\).
The algebraic procedure is a legitimate way of finding the inverse only after one has checked that the function is invertible, or after suitably fixing its domain and codomain. Solving an equation formally does not, by itself, guarantee the existence of an inverse function.
For example, from the relation
\[ y=x^2 \]
one formally obtains
\[ x=\pm\sqrt{y}. \]
This expression does not define an inverse function from \(\mathbb R\) to \(\mathbb R\), because a single positive value of \(y\) corresponds to two possible values of \(x\). The problem is not merely algebraic: the function \(f:\mathbb R\to\mathbb R\), \(f(x)=x^2\), is not injective and hence not invertible.
If instead we restrict the domain to \([0,+\infty)\) and consider
\[ f:[0,+\infty)\to[0,+\infty),\qquad f(x)=x^2, \]
then for every \(y\in[0,+\infty)\) there is a unique \(x\in[0,+\infty)\) such that \(x^2=y\), namely
\[ x=\sqrt{y}. \]
In this case the inverse is
\[ f^{-1}(x)=\sqrt{x}. \]
The computation of the inverse must therefore always be accompanied by a check of the domain, the codomain, and the bijectivity of the function.
Examples of invertible and non-invertible functions
To grasp the meaning of the inverse function better, it is useful to compare some examples in which invertibility depends crucially on the chosen domain and codomain.
An invertible function
Consider the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=x+4. \]
The function is injective, for if \(f(x_1)=f(x_2)\), then
\[ x_1+4=x_2+4, \]
and hence
\[ x_1=x_2. \]
It is also surjective, since for every \(y\in\mathbb R\) there is an \(x\in\mathbb R\) such that
\[ x+4=y. \]
Indeed it suffices to take
\[ x=y-4. \]
The function is therefore bijective and admits an inverse. From the relation
\[ y=x+4 \]
we obtain
\[ x=y-4. \]
Therefore
\[ f^{-1}(x)=x-4. \]
A function that is not invertible because it is not injective
Now consider the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=x^2. \]
This function is not injective. Indeed two distinct elements of the domain can have the same image:
\[ f(-2)=4 \qquad \text{and} \qquad f(2)=4. \]
If one tried to construct an inverse, the value \(4\) would have to be sent both to \(-2\) and to \(2\). This is impossible, since a function must assign a single value to each element of its domain.
Consequently, the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=x^2 \]
does not admit an inverse function.
A function that is not invertible because it is not surjective
Consider the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=e^x. \]
This function is injective, but it is not surjective onto \(\mathbb R\). Indeed its values are always positive:
\[ e^x>0 \qquad \text{for every } x\in\mathbb R. \]
Hence no real number less than or equal to zero belongs to the image of the function. For instance, there is no \(x\in\mathbb R\) such that
\[ e^x=-1. \]
Consequently, the function cannot have an inverse from \(\mathbb R\) to \(\mathbb R\), since any such inverse would have to be defined on the whole codomain \(\mathbb R\), including the values not attained by \(f\).
If, however, we change the codomain and consider
\[ f:\mathbb R\to(0,+\infty),\qquad f(x)=e^x, \]
then the function becomes bijective. In this case it admits an inverse
\[ f^{-1}:(0,+\infty)\to\mathbb R, \]
given by
\[ f^{-1}(x)=\ln x. \]
The same formula can give different functions
The previous examples illustrate an essential point: invertibility is not a property of the formula alone, but of the function as a whole.
The formula \(x^2\), regarded as a function from \(\mathbb R\) to \(\mathbb R\), does not define an invertible function. The same formula, regarded as a function from \([0,+\infty)\) to \([0,+\infty)\), does define an invertible one.
Likewise, the formula \(e^x\), regarded as a function from \(\mathbb R\) to \(\mathbb R\), is not surjective; regarded instead as a function from \(\mathbb R\) to \((0,+\infty)\), it becomes bijective.
To decide whether a function admits an inverse, then, three ingredients must always be specified: the rule of assignment, the domain, and the codomain.
Left inverse of a function
The inverse function exists, in the ordinary sense, only when the function is bijective. Nevertheless, even when a function fails to be bijective, part of the behavior of an inverse may still survive.
This leads to the notions of left inverse and right inverse.
Let
\[ f:A\to B \]
be a function. A function
\[ g:B\to A \]
is called a left inverse of \(f\) if
\[ g\circ f=\operatorname{id}_A. \]
Explicitly, this means that
\[ g(f(x))=x \qquad \text{for every } x\in A. \]
Thus, applying first \(f\) and then \(g\), we always return to the original element of the domain \(A\).
The name “left inverse” comes from the position of \(g\) in the composition
\[ g\circ f. \]
Indeed, in the expression \(g\circ f\), the function \(g\) appears to the left of \(f\). For this reason \(g\) is called a left inverse of \(f\).
The existence of a left inverse is closely tied to injectivity. If \(f\) has a left inverse, then \(f\) is injective.
Indeed, suppose there exist \(x_1,x_2\in A\) such that
\[ f(x_1)=f(x_2). \]
Applying \(g\) to both sides, we obtain
\[ g(f(x_1))=g(f(x_2)). \]
Since \(g\circ f=\operatorname{id}_A\), it follows that
\[ x_1=x_2. \]
Hence \(f\) is injective.
Conversely, if \(f\) is injective, then every element of the image of \(f\) comes from a unique element of \(A\). Hence, on those elements of \(B\) that lie in the image of \(f\), we can define a function carrying each value back to its unique starting point.
One important detail remains, however: if \(f\) is not surjective, some elements of \(B\) do not lie in the image of \(f\). On these elements the left inverse is not determined by the function \(f\), since they come from no element of \(A\).
For this reason, when \(f\) is injective but not surjective and \(A\) is nonempty, a left inverse can be defined in a natural way on the image of \(f\), whereas on the points of \(B\setminus f(A)\) its definition may be chosen arbitrarily, provided it takes values in \(A\).
Consider, for example, the function
\[ f:\mathbb R\to\mathbb R,\qquad f(x)=e^x. \]
This function is injective, but it is not surjective onto \(\mathbb R\), because its image is \((0,+\infty)\).
On the positive elements, that is, on the elements actually attained by \(f\), the function that inverts \(f\) is the natural logarithm:
\[ \ln(e^x)=x \qquad \text{for every } x\in\mathbb R. \]
Thus the function
\[ \ln:(0,+\infty)\to\mathbb R \]
inverts the exponential on its image, in the sense that
\[ \ln\circ \exp=\operatorname{id}_{\mathbb R}. \]
However, \(\ln\) is not, by itself, a left inverse of the function \(f:\mathbb R\to\mathbb R\), \(f(x)=e^x\), because it is not defined on the whole codomain \(\mathbb R\). To obtain a genuine left inverse \(g:\mathbb R\to\mathbb R\), one must extend the logarithm to the real values less than or equal to zero as well.
For example, we may define
\[ g(y)= \begin{cases} \ln y, & y>0,\\ 0, & y\le 0. \end{cases} \]
Then \(g:\mathbb R\to\mathbb R\) is well defined and, for every \(x\in\mathbb R\),
\[ g(e^x)=\ln(e^x)=x. \]
Hence
\[ g\circ \exp=\operatorname{id}_{\mathbb R}. \]
The choice of the value of \(g\) on the real numbers less than or equal to zero is arbitrary: at those points the function \(f(x)=e^x\) imposes no value.
The left inverse, therefore, guarantees that the function can be undone after it has been applied, but it does not require that every element of the codomain be attained.
In short, the existence of a left inverse expresses the fact that \(f\) does not identify distinct elements of the domain. For this reason the left inverse is linked to injectivity.
Right inverse of a function
Let
\[ f:A\to B \]
be a function. A function
\[ h:B\to A \]
is called a right inverse of \(f\) if
\[ f\circ h=\operatorname{id}_B. \]
Explicitly, this means that
\[ f(h(y))=y \qquad \text{for every } y\in B. \]
Thus, starting from an element \(y\in B\), the function \(h\) selects an element of \(A\) that \(f\) sends precisely to \(y\).
The name “right inverse” comes from the position of \(h\) in the composition
\[ f\circ h. \]
Indeed, in the expression \(f\circ h\), the function \(h\) appears to the right of \(f\). For this reason \(h\) is called a right inverse of \(f\).
The existence of a right inverse is closely tied to surjectivity. If \(f\) has a right inverse, then \(f\) is surjective.
Indeed, for every \(y\in B\), the identity
\[ f(h(y))=y \]
shows that \(y\) is the image of the element \(h(y)\in A\). Hence every element of \(B\) is attained by \(f\), and therefore \(f\) is surjective.
Conversely, if \(f\) is surjective, then for every \(y\in B\) there is at least one element \(x\in A\) such that
\[ f(x)=y. \]
To build a right inverse, for each \(y\in B\) we may choose one of the elements of \(A\) whose image is \(y\). Setting \(h(y)\) equal to one of these elements gives
\[ f(h(y))=y \qquad \text{for every } y\in B. \]
Thus \(h\) is a right inverse of \(f\).
In a general set-theoretic setting, this construction calls for a remark. To define \(h\) we must choose, for each \(y\in B\), an element of the fiber
\[ f^{-1}(\{y\})=\{x\in A : f(x)=y\}. \]
Since \(f\) is surjective, each of these fibers is nonempty. The simultaneous choice of one element from every fiber is, however, guaranteed in the general case by the axiom of choice. In the usual settings of elementary analysis and algebra this difficulty hardly ever arises, because the choices are normally explicit or fixed by a natural rule.
When \(f\) is surjective but not injective, the right inverse need not be unique. Indeed a single element \(y\in B\) may have several preimages in \(A\), and the function \(h\) must pick one of them.
Consider, for example, the function
\[ f:\mathbb R\to[0,+\infty),\qquad f(x)=x^2. \]
This function is surjective but not injective. Indeed every nonnegative real number is the square of at least one real number, yet if \(y>0\) then
\[ f(\sqrt y)=y \qquad \text{and} \qquad f(-\sqrt y)=y. \]
For each \(y\in[0,+\infty)\) we may take as preimage the nonnegative number \(\sqrt y\). This gives the function
\[ h:[0,+\infty)\to\mathbb R,\qquad h(y)=\sqrt y. \]
Then
\[ f(h(y))=f(\sqrt y)=(\sqrt y)^2=y \qquad \text{for every } y\in[0,+\infty). \]
Thus \(h\) is a right inverse of \(f\).
A different choice could, however, have been made, for instance
\[ k:[0,+\infty)\to\mathbb R,\qquad k(y)=-\sqrt y. \]
In this case as well
\[ f(k(y))=f(-\sqrt y)=(-\sqrt y)^2=y \qquad \text{for every } y\in[0,+\infty). \]
So \(k\) is a right inverse of \(f\) as well. This shows that, in the absence of injectivity, the right inverse may fail to be unique.
The right inverse, then, guarantees that every element of the codomain can be reached by suitably choosing an element of the domain. It does not, however, guarantee that such an element is unique.
In short, the existence of a right inverse expresses the fact that \(f\) reaches the whole codomain. For this reason the right inverse is linked to surjectivity.
The relationship between inverse, left inverse, and right inverse
The notions of left inverse and right inverse let us separate the two properties that, together, make a function invertible.
Let
\[ f:A\to B \]
be a function. A function
\[ g:B\to A \]
is a left inverse of \(f\) if
\[ g\circ f=\operatorname{id}_A. \]
This condition means that, after \(f\) has been applied, the function \(g\) makes it possible to return to the original element of \(A\). For this reason the existence of a left inverse is tied to the injectivity of \(f\).
A function
\[ h:B\to A \]
is instead a right inverse of \(f\) if
\[ f\circ h=\operatorname{id}_B. \]
This condition means that every element of \(B\) can be obtained by applying \(f\) to a suitable element of \(A\). For this reason the existence of a right inverse is tied to the surjectivity of \(f\).
When one and the same function
\[ u:B\to A \]
is simultaneously a left inverse and a right inverse of \(f\), that is, when both identities
\[ u\circ f=\operatorname{id}_A \]
and
\[ f\circ u=\operatorname{id}_B \]
hold, then \(u\) is the genuine inverse function of \(f\). In this case we write
\[ u=f^{-1}. \]
Thus the ordinary inverse function can be viewed as a function that is at once a left inverse and a right inverse.
In particular, if \(f\) admits an inverse function \(f^{-1}:B\to A\), then
\[ f^{-1}\circ f=\operatorname{id}_A \]
and
\[ f\circ f^{-1}=\operatorname{id}_B. \]
The first identity expresses injectivity: distinct elements of \(A\) are not identified by \(f\). The second identity expresses surjectivity: every element of \(B\) is attained by \(f\).
The situation can therefore be summarized as follows:
- the left inverse corresponds to recovering the elements of the domain after \(f\) has been applied;
- the right inverse corresponds to the possibility of representing every element of the codomain as an image under \(f\);
- the ordinary inverse function exists when both conditions are satisfied.
In terms of properties of the function:
\[ f \text{ has a left inverse } \Longrightarrow f \text{ is injective}, \]
while
\[ f \text{ has a right inverse } \Longrightarrow f \text{ is surjective}. \]
Conversely, if \(f\) is injective, then \(f\) can be inverted on its image; if \(f\) is surjective, then, assuming the axiom of choice in the general case, a right inverse can be constructed.
When \(f\) is both injective and surjective, the two conditions merge: the function admits a single inverse, which is at once a left inverse and a right inverse.
In symbols:
\[ f \text{ is bijective} \quad \Longleftrightarrow \quad \exists\, f^{-1}:B\to A \]
such that
\[ f^{-1}\circ f=\operatorname{id}_A \qquad \text{and} \qquad f\circ f^{-1}=\operatorname{id}_B. \]
This formulation brings out the deeper meaning of invertibility: a function is invertible when it loses no information about the elements of the domain and, at the same time, reaches every element of the codomain.
Summary
The inverse function lets us traverse a function in the opposite direction. If a function
\[ f:A\to B \]
assigns to an element \(x\in A\) the value \(y=f(x)\in B\), then the inverse function, when it exists, assigns to \(y\) the element \(x\) from which it came.
For this to be possible on the whole codomain \(B\), every element of \(B\) must be the image of one and only one element of \(A\). The existence requirement is surjectivity; the uniqueness requirement is injectivity.
Hence a function admits an inverse if and only if it is bijective:
\[ f:A\to B \text{ admits an inverse } f^{-1}:B\to A \quad \Longleftrightarrow \quad f \text{ is bijective}. \]
When the inverse exists, it is characterized by the two identities
\[ f^{-1}\circ f=\operatorname{id}_A \]
and
\[ f\circ f^{-1}=\operatorname{id}_B. \]
The first identity says that, starting from an element of \(A\) and applying first \(f\) and then \(f^{-1}\), we return to the original element. The second says that, starting from an element of \(B\) and applying first \(f^{-1}\) and then \(f\), we return to the original element.
The left and right inverses separate these two conditions.
A left inverse of \(f\) is a function \(g:B\to A\) such that
\[ g\circ f=\operatorname{id}_A. \]
It makes it possible to recover every element of the domain after \(f\) has been applied. For this reason it is linked to injectivity.
A right inverse of \(f\) is a function \(h:B\to A\) such that
\[ f\circ h=\operatorname{id}_B. \]
It makes it possible to obtain every element of the codomain as an image under \(f\). For this reason it is linked to surjectivity.
If one and the same function is simultaneously a left inverse and a right inverse of \(f\), then it is the ordinary inverse function of \(f\).
In conclusion:
- injectivity prevents two distinct elements of the domain from having the same image;
- surjectivity guarantees that every element of the codomain is attained;
- bijectivity guarantees both properties and makes the inverse function possible;
- the left inverse reflects injectivity;
- the right inverse reflects surjectivity;
- the ordinary inverse function exists when both conditions hold together.
The notion of inverse function, then, does not depend on a formula alone, but on the function considered in its entirety: domain, codomain, and rule of assignment. Only by specifying all these ingredients can one decide precisely whether a function is invertible and what its inverse is.