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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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    What do you mean by $g(x) > f(g(x))$?2012-11-04
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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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    i am really confused right now. is there a really easy and intuitive way to see this?2012-11-04
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    Not sure what you are confused about, and what is the "this" you want to see in an easy way. Functions between sets of one, two, and three elements are the easiest you can get.2012-11-04
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    what you just said. i mean can you just show me a diagram or something really simple.2012-11-04
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    f o g is onto and one-to-one if g is onto and f is one-to-one?2012-11-04
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    No, the first example in my answer shows an example where $g$ is onto, $f$ is one-to-one, and $f\circ g$ is neither onto nor one-to-one.2012-11-04
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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