Injective means one-one.
If $f$ and $g$ are injective then $gof$ must be injective. So, S1 is TRUE.
$gof$ is surjective (onto) meaning - codomain = range. So, $gof$ maps every element in C which is its codomain set. Since $gof$ is surjective $g$ must be surjective. Now $g$ is an injection meaning it maps one-one. Thus, $g$ becomes bijection.
$g$ being bijective means, for every element in B there is a corresponding unique element in C and vice-versa. $gof$ maps to every element in C. So, it must be the case that $f$ maps to every element in B- otherwise the corresponding element in C won't be mapped to by $g$. Hence, $f$ must be surjective.
So, C option is correct.
