Class on 06-10-2026

Today we mainly talked about injectivity/surjectivity of functions.

Composition of functions

In Exercise 3 we needed to think about injectivity/surjectivity of compositions.

Proof: Let $x \neq y$ be arbitrary elements of $A.$ Because $f$ is injective, $f(x) \neq f(y).$ Now, notice that $f(x)$ and $f(y)$ are two non-equal elements of the set $B.$ Using that $g$ is injective, $g(f(x)) \neq g(f(y)).$ Since $g(f(x)) = (g \circ f)(x)$ by definition, we have proven that for any $x \neq y,$ $$(g \circ f)(x) \neq (g \circ f)(y).$$ This means that $g \circ f$ is injective.

Remarks

At the start of the class I posed the question: if $$f \colon A \to A$$ is an injective function, is $f$ necessarily surjective? Answer: this statement is true for finite sets, but not infinite ones.

One exercise asked us to prove that a strictly increasing function $f \colon \R \to \R$ is necessarily injective. Indeed, if $x \neq y,$ then either $x > y$ or $y > x.$ In the first case $f(x) > f(y),$ and in the second case $f(y) > f(x).$ In either case, $f(x) \neq f(y).$ (We could also say: "Suppose without loss of generality that $x > y$…")

Then I asked a sort of converse: is an injective function $f \colon \R \to \R$ necessarily strictly increasing or decreasing? I argued that we can't make a counterexample by drawing a graph without lifting our pen. Someone found a counterexample, which was something of the form $$f(x) = \begin{cases} x & : x \not \in [0, 1] \\ 1 - x & : x \in [0, 1]. \end{cases}$$ (Actually in class I got the example subtly wrong; instead of $[0, 1]$ I wrote $[0, 1).$ That doesn't work, because in that case we would have $f(0) = f(1) = 1$…)