Inverses of Functions

What does it mean to "undo" a function? If $f$ takes $a$ to $b$, the inverse should take $b$ back to $a$. For that to be a well-formed function we need two things: every $b$ in the codomain must have some preimage (so $f^{-1}(b)$ is defined for all $b$) and only one preimage (so $f^{-1}(b)$ is unambiguous). Those are exactly surjectivity and injectivity. Bijection is the inevitable precondition for invertibility. Once you have an inverse, you have a powerful symmetry: encryption and decryption, encoding and decoding, undo and redo, Fourier transform and its inverse — all of these are bijection-and-inverse pairs in disguise.

The inverse function

Let $f: A \to B$ be a bijection. Then there is a unique function $f^{-1}: B \to A$, called the inverse of $f$, satisfying

Equivalently: $f^{-1} \circ f = \text{id}_A$ and $f \circ f^{-1} = \text{id}_B$. The inverse exists if and only if $f$ is bijective:

• If $f$ is not injective, two preimages would compete and $f^{-1}$ would not be well-defined.
• If $f$ is not surjective, some $b$ would have no preimage and $f^{-1}(b)$ would be undefined.

Finding inverses (the formula method)

To compute $f^{-1}$ when $f$ is given by a formula:

• Write $y = f(x)$.
• Solve for $x$ in terms of $y$.
• The resulting expression is $f^{-1}(y)$.
• Verify: check $f(f^{-1}(y)) = y$ and $f^{-1}(f(x)) = x$.

Example: $f(x) = 3x + 7$. Set $y = 3x + 7$, solve: $x = (y - 7)/3$. So $f^{-1}(y) = (y - 7)/3$. Verify: $f(f^{-1}(y)) = 3 \cdot \tfrac{y-7}{3} + 7 = y$. $\checkmark$

Inverse of a composition

If $f: A \to B$ and $g: B \to C$ are bijections, then $g \circ f$ is a bijection and

Note the order reversal: to undo "first $f$, then $g$," you must first undo $g$, then undo $f$ — like reversing the order in which you put on socks and shoes.

Pause and think: Why does $f(x) = x^2$ on $\mathbb{R} \to \mathbb{R}$ have no inverse? But if we restrict to $f: [0, \infty) \to [0, \infty)$, an inverse exists — what is it? Which obstacle did the restriction remove?

Try it

• Compute the inverse of $f(x) = 5x - 2$ and verify by composing in both directions.
• Predict before computing: what is $f^{-1}$ for $f(x) = (x - 3)/2$?
• Let $f(x) = 4x + 1$ and $g(x) = 2x - 3$. Use the rule $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ to compute $(g \circ f)^{-1}$ two ways and check they agree.
• Show: if $f$ is injective but not surjective, then $f$ has a left inverse (a function $g$ with $g \circ f = \text{id}_A$) but no two-sided inverse.
• Prove $(f^{-1})^{-1} = f$.

A trap to watch for

The notation $f^{-1}$ is overloaded. For a function it can mean:

• The inverse function, which requires $f$ to be bijective.
• The preimage of a set: $f^{-1}(T) = {a \mid f(a) \in T}$, which is defined for every function (no bijection needed). We cover this in §5.5.

And in calculus, $f^{-1}$ is sometimes written for the reciprocal $1/f$. Always check from context which meaning is intended.

What you now know

You can decide whether a function is invertible, compute its inverse by the formula method, verify both compositional identities, and apply the order-reversal rule for inverses of compositions. Section 5.4 turns to a different but related construction: the closure of a set under an operation, which is the smallest extension that "respects" the operation.

References

• Velleman, D. J. (2019). How to Prove It (3rd ed.). Cambridge University Press, §5.3.
• Hammack, R. (2018). Book of Proof, ch. 12.
• Halmos, P. R. (1960). Naive Set Theory. Springer, §10.
• Katz, J., Lindell, Y. (2014). Introduction to Modern Cryptography (2nd ed.). CRC Press, ch. 3 (block ciphers as keyed bijections).
• Cooley, J. W., Tukey, J. W. (1965). "An algorithm for the machine calculation of complex Fourier series." Math. Comp. 19, 297-301.