Let S- T- U be three non-void sets and f- S - T- g- T - U be so that g - f- S - U is surjective- Then

The correct option is B g is surjective, f may not be sog ∘ f : S → U is surjective ⇒∀ u∈ U ∃ s ∈ S such that g(f(s))=u ⇒∀ u∈ U ∃ an element t∈ T such that g(t ...

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Byju's Answer

Standard XII

Chemistry

Arrhenius Acid

Let S, T, U b...

Question

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

g

and

f

are both surjective

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g

is surjective,

f

may not be so

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f

is surjective,

g

may not be so

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f

and

g

both may not be surjective

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Solution

The correct option is B $g$ is surjective, $f$ may not be so
$g \circ f : S \to U$ is surjective
$\Rightarrow \forall u \in U \exists s \in S$ such that $g (f (s)) = u$
$\Rightarrow \forall u \in U \exists$ an element $t \in T$ such that $g (t) = u$
$\Rightarrow g$ is surjective.
But there is no need of $f$ to be surjective. Example:

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Let S,T,U be three non-void sets and f:S→T,g:T→U be so that g∘f:S→U is surjective. Then

The correct option is B g is surjective, f may not be sog∘f:S→U is surjective ⇒∀ u∈U ∃ s∈S such that g(f(s))=u ⇒∀ u∈U ∃ an element t∈T such that g(t)=u ⇒g is surjective. But there is no need of f to be surjective. Example:

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