5
$\begingroup$

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?

$\endgroup$

closed as off-topic by Najib Idrissi, N. F. Taussig, Empty, quid, Hagen von Eitzen Oct 25 '15 at 22:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – N. F. Taussig, Empty, Hagen von Eitzen
If this question can be reworded to fit the rules in the help center, please edit the question.

9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How does this diagram show that the first function is 1:1 and the second function onto? $\endgroup$ – johnnymath Oct 26 '11 at 1:15
  • $\begingroup$ @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. $\endgroup$ – Arturo Magidin Oct 26 '11 at 1:50
6
$\begingroup$

HINT: $$\begin{array}{}&&\bullet&&\\ &&&\searrow&\\ \bullet&\to&\bullet&\to&\bullet\\ X&f&Y&g&X \end{array}$$

$\endgroup$
  • $\begingroup$ You and Arturo with the exact same diagram? (Well, modulo labels...) $\endgroup$ – Asaf Karagila Oct 25 '11 at 21:24
  • 3
    $\begingroup$ @Asaf: Mine’s prettier $-$ more symmetric. :-) $\endgroup$ – Brian M. Scott Oct 25 '11 at 21:26
4
$\begingroup$

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:

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.