# If $g \circ f$ is the identity function, then which of $f$ and $g$ is onto and which is one-to-one? [closed]

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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## closed as off-topic by Najib Idrissi, N. F. Taussig, S.Panja-1729, quid, Hagen von EitzenOct 25 '15 at 22:42

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The question is not that difficult. Any guesses from your side? – Srivatsan Oct 25 '11 at 20:58
Possible duplicate of Show that if $g \circ f$ is injective, then so is $f$. – Najib Idrissi Oct 25 '15 at 16:17

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

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
@Asaf: Mine’s prettier $-$ more symmetric. :-) – Brian M. Scott Oct 25 '11 at 21:26

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: