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Let there be two arbitrary functions $f$ and $g$, and let $g \circ f$ be their composition.

  1. Suppose $g \circ f$ is one-one. Then prove that this implies $f$ is also one-one.

  2. Suppose $g \circ f$ is onto. Then prove that this implies $g$ is also onto.

I tried to do both of the proofs by contradiction method but got stuck in that approach. Please help me in this proof. Proofs in more than one way is welcomed.

Patrick Stevens
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3 Answers3

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SUPPOSE $x,y\in A$ such that $f(x)=f(y)$. Then $g(f(x)) = g(f(y))$. But this is the same as $(g\circ f)(x) = (g\circ f)(y)$ and $g\circ f$ being injective $\implies x=y$. This shows that $f$ is injective.

Fawad
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I just prove (1), the other is similar. If $g\circ f$ is injective, then $g(f(x))=g(f(y))$ iff $x=y$. The $f$ has to be injective too, indeed, in the oppositive, there would exist $x<y$ such that $f(x)=f(y)$ and this would imply $g(f(x))=g(f(y))$.

Paolo Leonetti
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Take $f:A\to B$ and $g:B\to C$.

$(a)$ Let $x_1, x_2 \in A$ and suppose that $f(x_1) = f(x_2)$. Then $(g\circ f)(x_1) = g(f(x_1)) = g(f(x_2)) = (g\circ f)(x_2)$. But since $g\circ f$ is injective, this implies that $x_1 = x_2$. Therefore $f$ is injective.

$(b)$ Suppose that $g\circ f$ is surjective. Let $z \in C$. Then since $g\circ f$ is surjective, there exists $x \in A$ such that $(g\circ f)(x) = g(f(x)) = z$. Therefore if we let $y = f(x) \in B$, then $g(y) = z$. Thus $g$ is surjective. Hope it helps.