Is it true that $"\exists f: A\to B, \ f\text{ bijective}" \iff "\exists f: B\to A, \ f\text{ bijective}"$ Is it true that 
$$"\exists f: A\to B, \quad f\quad bijective" \iff "\exists f: B\to A, \quad f\quad bijective"$$
Is it the same $f$? 
Intuitively I would say yes to both of these questions, but this example got me doubting: 
Let $f(x)=e^x.$ It is true that $"f:\mathbb{R}\to\mathbb{R}^+, \quad f\quad bijective"$, since every element in $\mathbb{R^+}$ can be reached (surjective) and from simply looking at the graph follows that it is injective. 
However, $"f:\mathbb{R^+}\to \mathbb{R}, \quad f \quad bijective"$ does not seem true since we can't reach the negative elements of $\mathbb{R}$ (not surjective).
What's going on? 
 A: You need to take the inverse: if $f:A\to B$ is a bijection, then really it is the function $f^{-1}$ that goes backwards from $B$ to $A$. In your example, $f$ is the exponential function, and the inverse is $\log: \mathbb{R}^+ \to \mathbb{R}$. 
Hopefully this convinces you that your first question is yes: there is a bijection $f:A\to B$ if and only if there is one from $B\to A$; it just won't be the same function.
A: Yes.  $f:A\rightarrow B$ is bijective means that $f$ is one-one and onto.  Another way to say this is that $f$ is an invertible function.  It's inverse is the bijection that you're looking for.  But let's prove this.
Does there exist a $g:B\rightarrow A$ such that $g$ is one-one?  $f$ is one-one means that for all $y$ in $\text{Image}(f)$, there exists exactly one $x$ such that $f(x)=y$.  Let $g$ be the mapping such that $g(y) = g(f(x)) = x$.  Then $g$ is one-one.
Is $g$ onto?  $f$ is onto means that for all $y$ in $\text{Image}(f)$ there exists an $x$ such that $f(x)=y$.  Since $g$ is the inverse mapping of $f$, $\text{Image}(g)=\text{Domain}(f)$.  Therefore $g$ is onto.
And thus $g$ is a bijection.
A: It's called the inverse function.
Let $f: A\to B$ be a bijective function.
Let $f^{-1}: B\to A$ be defined as.
For $x \in B$.  Let $y \in A$ be the element where $f(y) = x$.  Define $f^{-1}(x) = y$
We need to prove:
1) That that is a well defined function.  That is for every $x\in B$ then there actually does exist a $y\in A$ so that value so that $f^{-1}(x) = y$.  And that such a $y$ is unambiguous.  There is only one unique value of $y \in A$ that can possible work so that $f^{-1}(x) = y$ will work.
And 2) We have to prove it is a bijection.
Pf of 1: As $f$ in surjective then for every $x \in B$ there is a $y \in A$ so that $f(y) = x$.  And because $f$ is injective there is only one unique $y \in A$ so that $f(y) = x$.
That's it.  $f^{-1}(x) = $ the $y$ so that $f(y) = x$ is well defined.
Pf of 2:
a) Pf: that $f^{-1}$ is surjective.  Let $w  \in A$.  Then $f(w) \in B$.  Let $x = f(w)$.  Then $f^{-1}(x) = w$. So $f^{-1}$ is surjective.
b) Pf: the $f^{-1}$ is injective.  Suppose $f^{-1}(x) = f^{-1}(w)=y$ for $x,w \in B$ and $y \in A$.  Then $f(y) = x$ and $f(y) = w$.  So $x=w$.  So $f^{-1}$ is injective.
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does not seem true since we can't reach the negative elements of R (not surjective).

But you can!
If $x < 0$ then $f(x) = e^x = y$ is such that $0 < e^x = y < 1$.
So $f^{-1}(y) = x$ and $x < 0$.  That's a negative number.
In fact if $f(x) = e^x$ then $f^{-1}(y) = \ln y$.  And as you know.  If $0 < y < 1$ then $\ln y < 0$.  And if $y \ge 1$ then $\ln y \ge 0$.
And $\ln: \mathbb R^+ \to \mathbb R$ is indeed surjective. (and injective).
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Ah!  I see the problem is that the statement used the the same term $f$ in the first clause as the second. 
This is acceptable (but not advisable) because the $\exists$ clauses are separate.   
In casual english the statement reads:
With two sets $A$ to $B$,
Condition 1: There is a bijective function, lets call it $f$, from $A$ to  $B$.  
Condition 2: There is a bijective function, lets call it $f$, from $B$ to $A$.
Condition 1 if and only if Condition 2.
The functions both referred to as $f$ do no both refer to the same function.  Indeed if $A$ and $B$ are different sets this is would clearly be false if the two functions were the same they have different domains.
Technically the domain is part of the definition of a function and if functions have different names they are different functions.
But if you take the naive (and in my opinion, reasonable) mistake that a function is the rule and can be restricted and extended to different domains and ranges and still be the same function this is still obviously false.  Just pick an $x$ where $f(x) = y;$ but $f(y)$ either doesn't exist or $f(y) \ne x$ and then you have $f:\{x\} \to \{y\}$ is a bijection but $f:\{y\}\to \{x\}$ isn't even a reasonable function.
If $A,B$ aren't subsets of the same "space" this concept is pure gibberish.   Let $f:\{"a", "b", .... ,"z"\} \to \mathbb N_{26}$ via $f(a) = 1; f(b) = 2$ etc. but $f:\mathbb N_{26}\to \{"a", "b", .... ,"z"\}$ is nonsense.  $f(n)$ isn't defined.  $f$ takes letters as its input; not numbers.
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