Lecture 4: proofs and functions


Definitions: let \(f : A → B\). Then \(f\) is


Definition: if \(f : A → B\) and \(g : B → C\) then the composition of g with f (written \(g \circ f\)) is the function \((g \circ f) : A → C\) is given by \((g \circ f)(x) = g(f(x))\).

Note that \(g \circ f\) is different from \(f \circ g\).


Definition: The identity function on X is the function \(id : X → X\) given by \(id(x) = x\).

Definition: If \(f : A → B\) then \(g\) is a left inverse of \(f\) if \(g \circ f = id\). In other words, if for all \(x \in A\), \(g(f(x)) = x\) (Note: \(g\) is a left inverse because you write it on the left).

Similarly, \(g\) is a right inverse if \(f \circ g = id\).

'jectivity and inverses

Claim: \(f\) has a right inverse if and only if it is surjective.

Proof: We must prove that if \(f\) has a right inverse, then it is surjective, and also that if \(f\) is surjective, then it has a right inverse.

(RI ⇒ surj) Suppose that \(f\) has a right inverse \(g\). We will show that \(f\) is surjective by contradiction. Suppose that \(f\) is not surjective. Then there exists some \(y\) such that for all \(x\), \(f(x) \neq y\) (). Now, since \(g\) is a right inverse of \(x\), \(f(g(y)) = y\). But this contradicts (), so our original assumption (that \(f\) is not surjective) must be false.

(surj ⇒ RI) Suppose \(f\) is surjective. We will show that \(f\) has a right inverse by constructing it. Let \(g : B → A\) be defined as follows. Given any \(y \in B\) we know that there exists some \(x \in A\) such that \(f(x) = y\) (since \(f\) is surjective). Choose one of them, and define \(g(y)\) to be \(x\).

I claim \(g\) is a right inverse of \(f\). Indeed, for any \(y \in B\), we have \(f(g(y)) = y\) by the definition of \(g\).

This concludes the proof.

Note: it turns out that \(f\) is injective if and only if it has a left inverse, and \(f\) is bijective if and only if it has a two-sided inverse. This will be on the next homework.