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Say $f:X\rightarrow Y$ and $g:Y\rightarrow X$ are functions where $g\circ f:X\rightarrow X$ is the identity. Which of $f$ and $g$ is onto, and which is one-to-one?

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The question is not that difficult. Any guesses from your side? –  Srivatsan Oct 25 '11 at 20:58
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@johnny: In the past day you have asked four questions of a similar homework-y nature. math.SE is not a place to get your homework solved. If there are concepts which confuse you, feel free to ask about them. Mathematicians would usually go pretty far to help someone understand, but homework assignments are given for a reason. If you don't get your hands dirty with proofs, it will only harm you later on. Lastly, if this is indeed a homework assignment, use the [homework] tag. –  Asaf Karagila Oct 25 '11 at 21:17
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4 Answers

up vote 8 down vote accepted

If it is just a matter of remembering what the right conclusion is, here's the picture I always use to remember: $$\begin{array}{rcl} &\bullet &\\ &&\searrow\\ \bullet\rightarrow& \bullet & \rightarrow\bullet\\ X\quad\quad&Y&\quad\quad Z \end{array}$$ The compositum is one-to-one and onto: the first function is one-to-one but not onto; the second function is onto, but not one-to-one.

So: if a compositum is one-to-one, the first function applied is one-to-one. If a compositum is onto, then the second function applied is onto.

If $g\circ f = \mathrm{id}$, then the first function ($f$) is one-to-one, and the second function ($g$), is onto.

If it is a matter of proving that the first function is one-to-one and the second function is onto, well, you'd need a proof. An example does not suffice.

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How does this diagram show that the first function is 1:1 and the second function onto? –  johnnymath Oct 26 '11 at 1:15
    
@johnnymath: Look at the first function, the one that goes from $X$ to $Y$. Is it one-to-one? Yes. Is it onto? No. Look at the second function in the diagram, the one that goes from $Y$ to $Z$. Is it one-to-one? No. Is it onto? Yes. That's how. –  Arturo Magidin Oct 26 '11 at 1:50
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HINT: $$\begin{array}{}&&\bullet&&\\ &&&\searrow&\\ \bullet&\to&\bullet&\to&\bullet\\ X&f&Y&g&X \end{array}$$

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You and Arturo with the exact same diagram? (Well, modulo labels...) –  Asaf Karagila Oct 25 '11 at 21:24
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@Asaf: Mine’s prettier $-$ more symmetric. :-) –  Brian M. Scott Oct 25 '11 at 21:26
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I think $f$ should be one-to-one and $g$ should be onto, since $g$ has to cover all of $X$ in its range and $f$ has to make $X$ correspond in a one-to-one fashion with $Y$. It seems that $g$ could be not one-to-one if it is an inverse of $f$ that discards some of the information that being a member of Y conveys. For example, if $X = \mathbb{R}^{+}$, $Y = \mathbb{R}$, $f(x) = x$, $g(x) = |x|$, $g$ is not one-to-one, but it is onto, and f is one-to-one, but clearly not onto. Hopefully this example is valid and helps you out.

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You already have several answers which can help you remember the theorem. If you're looking for a proof (and have problems with showing it yourself), you might try to have a look at these links:

or these questions/answers:

More general results are here:

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