(related to Proposition: Composition of Injective Functions is Injective)
- Assume, for some $a_1,a_2\in A$, we have $(g\circ f)(a_1)=(g\circ f)(a_2)$.
- By the definition of composition, $g(f(a_1))=g(f(a_2)).$
- Since $g:B\to C$ is injective, it follows $f(a_1)=f(a_2).$
- Since $f:A\to B$ is injective, it follows that $a_1=a_2$.
- Altogether, we have concluded from $(g\circ f)(a_1)=(g\circ f)(a_2)$ in $C$ that $a_1=a_2$ for any $a_1,a_2$ in $A$.
- Thus, $(g\circ f)$ is injective.
- Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016