Let $f$ be a function from $A$ to $B$ and $g$ a function from $B$ to $C$. Show that if the composite function $g \circ f$ is one–to–one (injective), then $f$ is one–to–one (injective).
My understanding is that, for infectivity, we have to prove both
- If $a_1 = a_2$, then $f(a_1) = f(a_2)$, and
- If $f(a_1) = f(a_2)$, then $a_1 = a_2$.
My proof was going to go something like this.
Proof
Suppose $g \circ f$ is injective (one-to-one), where $f: A \to B$ and $g: B \to C$.
(Part 1) Let $f(x_1) = f(x_2)$, where $x_1, x_2 \in A$.
Therefore, $g(f(x_1)) = g(f(x_2))$. But since $g \circ f$ is injective, each element in the image of $g \circ f$ is mapped to by only one element in the domain of $g \circ f$.
Therefore, we must have that $x_1 = x_2$.
I was then going to do part 2, which would instead assume that $x_1 = x_2$, where $x_1 = x_2 \in A$. But the proof that I saw online did not do this part; instead, they only did part 1 and said that the proof is complete. Wouldn't we have to do both parts, since it is a logical equivalence (injectivity means $a_1 = a_2 \Longleftrightarrow f(a_1) = f(a_2)$)?
And if so, can someone please demonstrate the second part?
Let $r,t \in A$. Suppose $f(r)=f(t)$, then $g(f(r))=g(f(t))$. But $g \circ f$ is injective implies $r=t$. Thus $f$ is injective.