Solution
= Solution
Suppose $(g\circ f)(a)=(g\circ f)(a')$. <Injectivity> of $g$ gives $f(a)=f(a')$, and <injectivity> of $f$ then gives $a=a'$. Thus the <function composition> is <injective>.
= Solution
Suppose $(g\circ f)(a)=(g\circ f)(a')$. <Injectivity> of $g$ gives $f(a)=f(a')$, and <injectivity> of $f$ then gives $a=a'$. Thus the <function composition> is <injective>.