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

enter image description here

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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