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Lecture 13: inverse functions

• Reading: MCS 4.3-4.5

• definitions: composition, identity function, left inverse, right inverse, two sided inverse
• theorems
  • $f$ is injective if and only if it has a left inverse
  • $f$ is surjective if and only if it has a right inverse
  • $f$ is bijective if and only if it has a two-sided inverse

Definitions

If $f : A → B$ and $g : B → C$ then the composition of $g$ and $f$ (written $g \circ f$)$ is the function $g \circ f : A → C$ given by $(g \circ f)(a) := g(f(a))$.

Note: pay attention to the domains and codomains; with $f$ and $g$ as given, $f \circ g$ does not make sense, because $g(b) ∈ C$ so $f(g(b))$ is not defined.

The identity function on a set $A$ is the function $id_A : A → A$ given by $id_A(x) := x$. We often omit $A$ when it is clear from context.

If $f : A → B$ and $g : B → A$, and $g \circ f = id_A$ then we say $f$ is a right-inverse of $g$ and $g$ is a left-inverse of $f$. The left- and right- refer to which side of the $\circ$ the function goes; $g$ is a left-inverse of $f$ because when you write it on the left of $f$, you get the identity.

Theorems

The following claims are true:

• $f : A → B$ is injective if and only if it has a left-inverse
• $f : A → B$ is surjective if and only if it has a right-inverse
• $f : A → B$ is bijective if and only if it has a two-sided inverse

Note: The way to remember (and prove) these is to draw yourself a picture of an injection (or surjection), draw the best inverse you can, and then see which way the composition works. If I don't draw a picture, I easily get left and right mixed up.

These are all good proofs to do as exercises. We did the first of them in class:

Claim: if $f : A → B$ is injective and $A ≠ \emptyset$, then $f$ has a left-inverse.

Proof: Suppose $f : A → B$ is injective. Note that since $A \neq \emptyset$, there exists some $a_0 \in A$. Let $g : B → A$ be defined as follows. Given $b \in B$, if $b = f(a)$ for some $a$ in $A$, then let $g(b) := a$. If $b$ is not in the image of $f$, then define $g(b) := a_0$.

I claim $g$ is a left-inverse of $f$. In other words, $g \circ f = id$. In other words, $∀ a ∈ A$, $g(f(a)) = a$. To see this, choose an arbitrary $a \in A$. Then $f(a)$ is in the image of $f$, so by definition of $g$, we have $g(f(a)) = a'$ for some $a'$ satisfying $f(a') = f(a)$. But since $f$ is injective, we know $a' = a$, which is what we wanted to prove.

Note: While writing this proof, it helps to draw yourself a picture of a simple injective function, and think about how you would construct the inverse.

Claim: Suppose $f : A → B$ has a left-inverse. Then $f$ is injective.

Proof: Let $g$ be a left inverse of $f$. Then $g \circ f = id$. We want to show that $f$ is injective, i.e. for all $a_1, a_2 \in A$, if $f(a_1) = f(a_2)$ then $a_1 = a_2$. Choose arbitrary $a_1$ and $a_2$ and assume that $f(a_1) = f(a_2)$. Applying $g$ to both sides of the equation gives $g(f(a_1)) = g(f(a_2))$. But $g(f(a_1)) = a_1$ (and likewise for $a_2$) so $a_1 = a_2$.

The proofs of the remaining claims are mostly straightforward and are left as exercises. The proof of one direction of the third claim is a bit tricky:

Claim: If $f : A → B$ is bijective, then it has a two-sided inverse.

Proof: Since $f$ is bijective, by the previous claims we know it has a left inverse $g_l : B → A$ and a right inverse $g_r : B → A$. I will show that $g_l = g_r$, which means that $g_l$ is a two-sided inverse.

To see this, choose an arbitrary $b \in B$. We want to show that $g_l(b) = g_r(b)$. Consider $g_l(f(g_r(b))$. Since $g_l \circ f = id$, we have $g_l(f(g_r(b)) = g_r(b)$. On the other hand, since $f \circ g_r = id$, we have $g_l(f(g_r(b)) = g_l(b)$. Combining these shows $g_r(b) = g_l(b)$ as required.