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If $g$ is one-to-one and $f$ is onto, then we can't say anything about $f \circ g$, correct?

and if $f\circ g$ is one-to-one and onto, then $g$ is one-to-one and $f$ is onto?

$g(x) > f(g(x)) = f \circ g$

$A > Z$ and $Z > B$

we need to look at set $A$ (all elements of $A$ have to map to one unique element of $Z$ to see whether a function $g$ is one-to-one and set $B$ to see if $f$ is onto (there's an arrow to every elements).

Just wanted to make sure I understood correctly.

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    its an arrow one function to the other2012-11-04

1 Answers 1

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First, if $f\circ g$ is one-to-one, then $g$ is one-to-one. And if $f\circ g$ is onto, then $f$ is onto. So the assertion "if $g$ is onto and $f$ is one-to-one then $f\circ g$ is onto and one-to-one" has to be false.

For example, let $A=\{1,2\}$, $B=\{1\}$, $C=\{1,2\}$. Let $g:A\to B$ be $g(1)=g(2)=1$, $f:B\to C$ be $f(1)=1$. Then $g$ is onto, $f$ is one-to-one, but $f\circ g$ is neither.

In the other way, as you say, you cannot say anything either. Let $A=\{1,2\}$, $B=\{1,2,3\}$, $C=\{1,2\}$. Let $g:A\to B$ be $g(1)=1$, $g(2)=2$, so it is one-to-one. Let $f:B\to C$ be $f(1)=f(2)=1$, $f(3)=2$. Then $f$ is onto. And $f\circ g:A\to C$ is $f\circ g(1)=f\circ g(2)=1$, so it is neither one-to-one nor onto.

I have to admit I struggle to understand the rest of your post.

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    The only implications that hold in the direction you are looking for are that if **both** $f$ and $g$ are one-to-one, then $f\circ g$ is one-to-one; and if **both** $f$ and $g$ are onto, then $f\circ g$ is onto.2012-11-04