Simple bijectiveness question If $f$'s image set is $g$'s domain and vice versa, does that imply that their domains have a 1-1 correspondence? That $f$ and $g$ are both bijective mappings? Are my questions even meaningful?
Edit:
My question stems from this example from David Brannan's First Course in Mathematical Analysis.
I don't feel entirely comfortable with his argument that f is a one-to-one correspondence. (I do know how to show this by showing surjectivity(which is assumed here) and injectivity separately.)

 A: No; consider $f:[-1,1]\to[0,1]$ given by $f(x)=x^2$ and $g:[0,1]\to[-1,1]$ defined by $g(x)=2x-1$. One of them is bijective in this case, but not both. 
As Jonas Meyer points out below, given AC, the domains will have a $1$-$1$ correspondence.
EDIT: The author's approach is certainly not how I would approach it; I'd calculate $$f'(x)=\frac{1}{(1-x)^2}>0\textrm{ for all }x\in(-\infty,1).$$However, as it stands, I think your revised question has best been answered by Andres; this is just an alternative method which I find more straightforward.
A: Call $D_f,D_g$, respectively, the domain of $f$, and the domain of $g$. Your assumption says that $f:D_f\to D_g$ is onto, and that $g:D_g\to D_f$ is onto. 
Now, assuming the axiom of choice, if there is a surjection $\pi:A\to B$, then there is an injection $\tau:B\to A$ (for each $b\in B$, we can choose one of its preimages in $A$). So, this gives us injections from $D_g$ into $D_f$ and viceversa.
The Bernstein-Cantor-Schroeder theorem ensures now that there is a bijection between $D_f$ and $D_g$.

On the other hand, $f$ and $g$ themselves do not need to be bijective, or injective. For example, we could have $f:(0,1)\to(0,1)$ be the function that multiplies a number by $10$ and discards the integer part, if any. So, for example, $f(.34)=f(.04)=.4$. Clearly, $f$ is onto but not $1$-$1$, and we can take $g=f$. 

Now: The use of choice is essential to get a positive answer to whether $D_f$ and $D_g$ are in bijection. Without choice, we can have sets $A$ and $B$ of incomparable size, and surjections in both directions. (We can even have both $A$ and $B$ to be quotients of $\mathbb R$ in an appropriate model of set theory where the full axiom of choice fails.) See this MO question for closely related issues, and links to examples, or see this blog post of mine. 

With regards to your example, the function $f$ is clearly defined on $(-\infty,1)$. It is also meaningful at other points, but we do not care. The function admits an inverse, which is obtained by solving $f(x)=y$ for $x$ in terms of $y$. That there is an inverse tells us that $f$ is injective (otherwise, for some values of $y$ we would not have been able to find a single $x$). Clearly, $f$ is onto the domain of this inverse function (by definition: The values of $y$ we are considering are precisely those for which $f(x)=y$ for some $x$). So what remains is to find the image of $f$. That's what's done at the end of the displayed image.
Of course, one could have gone at this differently, and show first that $f$ is injective by directly checking that $f(x_1)=f(x_2)$ gives us $x_1=x_2$. Then, we could find the range of $f$ very much as in the final 5 lines. Having identified the image $I$, by definition $f$ is onto $I$. Since it is injective, this directly shows that $f$ is a bijection between its domain and its image.
