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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6The question is not that difficult. Any guesses from your side? – Srivatsan Oct 25 '11 at 20:58
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1Possible duplicate of Show that if $g \circ f$ is injective, then so is $f$. – Najib Idrissi Oct 25 '15 at 16:17
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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.
Arturo Magidin
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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 01:15
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@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 01:50
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HINT: $$\begin{array}{}&&\bullet&&\\ &&&\searrow&\\ \bullet&\to&\bullet&\to&\bullet\\ X&f&Y&g&X \end{array}$$
Brian M. Scott
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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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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:
- http://www.proofwiki.org/wiki/Injection_if_Composite_is_an_Injection
- http://www.proofwiki.org/wiki/Surjection_if_Composite_is_a_Surjection
or these questions/answers:
- If g(f(x)) is one-to-one (injective) show f(x) is also one-to-one (given that...)
- Surjection on composed function?
- Injective and Surjective Functions
More general results are here:
Martin Sleziak
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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.
karmic_mishap
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