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Suppose the function $g$ and $f$ are one-to-one. Is $f \circ g$ one-to-one?

Suppose $f \circ g$ is one-to-one, are the function $g$ and $f$ one-to-one?

Suppose $f \circ g$ is onto, are the function $g$ and $f$ onto?

Suppose the function $g$ and $f$ are onto. Is $f \circ g$ onto?

I was trying to think of examples to respond to those questions, but I couldn't think of anything.

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2 Answers 2

HINTS: This picture should help you with the two that are not true.

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The two that are true are pretty easy to prove once you identify them. One of the true ones is the first one; just suppose that $(f\circ g)(x)=(f\circ g)(y)$, and use what you know about $f$ and $g$ to prove that $x=y$. For starters, what can you say about $g(x)$ and $g(y)$, if $(f\circ g)(x)=(f\circ g)(y)$?

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If you note that "$f\colon A\to B$ is injective" is equivalent to "There exists $h\colon B\to A$ such that $h\circ f=1_A$" and that "$f\colon A\to B$ is surjective" is equivalent to "There exists $h\colon B\to A$ such that $f\circ h=1_B$", the valid conclusions should be clear:

  • $f\colon B\to C,g\colon A\to B$ one-to-one $\Rightarrow$ $f\circ g\colon A\to C$ one-to-one.
  • $f\circ g$ one-to-one $\Rightarrow$ $f$ one-to-one.
  • $f\circ g$ onto $\Rightarrow$ $g$ onto. $x\mapsto x-1$.
  • $f,g$ onto $\Rightarrow$ $f\circ g$ onto.

Consider $f\colon \mathbb Z \to \mathbb N$, $x\mapsto |x|+1$ and $g\colon \mathbb N \to \mathbb Z$, $x\mapsto x-1$. Then $f\circ g$ is one-to-one and onto (in fact is the identity), but $f$ is not one-to-one and $g$ is not onto.

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