If $g \circ f$ is the identity function, then which of $f$ and $g$ is onto and which is one-to-one?  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?
 A: 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.
A: HINT: $$\begin{array}{}&&\bullet&&\\
&&&\searrow&\\
\bullet&\to&\bullet&\to&\bullet\\
X&f&Y&g&X
\end{array}$$
A: 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:


*

*http://www.proofwiki.org/wiki/Injection_iff_Left_Inverse

*http://www.proofwiki.org/wiki/Surjection_iff_Right_Inverse
A: 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.
