Solution
= Solution
Suppose $(g\circ f)(a)=(g\circ f)(a')$. The <injectivity> of $g$ implies $f(a)=f(a')$, and the <injectivity> of $f$ then implies $a=a'$. Hence $g\circ f$ is <injective>.
= Solution
Suppose $(g\circ f)(a)=(g\circ f)(a')$. The <injectivity> of $g$ implies $f(a)=f(a')$, and the <injectivity> of $f$ then implies $a=a'$. Hence $g\circ f$ is <injective>.