Let S, T, U be three non-void sets and f: S r | vedclass

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(B) We are given that $g \circ f: S \rightarrow U$ is a surjective (onto) function.By definition of a surjective function, for every element $z \in U$

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$g$ is surjective, $f$ may not be so

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(B) We are given that $g \circ f: S \rightarrow U$ is a surjective (onto) function.By definition of a surjective function, for every element $z \in U$, there exists at least one element $x \in S$ such that $(g \circ f)(x) = z$.This can be written as $g(f(x)) = z$.Let $y = f(x)$. Since $x \in S$ and $f: S \rightarrow T$, it follows that $y \in T$.Substituting this into the equation, we get $g(y) = z$.Since for every $z \in U$, we have found an element $y \in T$ such that $g(y) = z$, it follows that $g: T \rightarrow U$ is a surjective function.However, $f$ does not necessarily have to be surjective because the elements in $T$ that are not in the image of $f$ do not affect the surjectivity of $g \circ f$ as long as the image of $f$ covers enough elements in $T$ to map onto all of $U$ through $g$.

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ be functions such that $g \circ f: S \rightarrow U$ is surjective. Then,

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