Reading: MCS 4.3-4.5
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.
The following claims are true:
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.