Question: In Bonn, we've been have a discussion on the topic in the title: 

> 
> Suppose that A and B is are classes and that there are injections from A to B and fom B to A. Does it follow that there is a bijection between A and B?
> 
> 
> 

Example: Let A the class of sets of cardinality one and let B be the class of sets of cardinality two. There is an injection 

A -> B sending a to {a, emptyset},

B-> A sending b to {{b}}.

Does it follow that there is a bijection between A and B?

Answer: 
Since the question has arisen whether the standard
arguments really do work with classes, let me post this
answer giving some fuller details about one comparatively
concrete way to do it. Other methods are also possible.

I shall work in the theory of Goedel-Bernays GB set theory
(without global choice), a general setting for treating
classes, which forms a conservative extension of ZF. This
theory includes as a special case the traditional treatment
of classes as definable collections in ZF, since every
model of ZF, when augmented with its definable classes,
forms a model of GB. Thus, the GB context seems the most
comprehensive way to answer (and if you prefer ZF, then
just imagine that all classes here are definable from
parameters).

Suppose that $A$ and $B$ are classes and that we have class
functions $F:A\to B$ and $G:B\to A$ which are injective, as
in the question. Let $A\_0=A-\operatorname{ran}(G)$, which is a class,
since it is definable from $A$ and $G$. Let us say that a
sequence $\langle x\_0,x\_1,x\_2,\ldots\rangle$ is a
*back-and-forth-iteration sequence* if $x\_0\in A$ and
$x\_{2n+1}=F(x\_{2n})$ and $x\_{2n+2}=G(x\_{2n+1})$ for all natural numbers $n$. Let $A\_n$
be the elements $a\_{2n}$ that appear at the even
coordinates of a back-and-forth iteration sequence with
$a\_0\in A\_0$. This notion is definable from $F$ and $G$, and the point is that we have a uniform presentation of the $A\_n$ in a single class $\{(n,a)\mid a\in A\_n\}$.
Let $A^+=\bigcup\_n A\_n$, which is a class definable from
$F$ and $G$.

Let $H$ be the function $(F\upharpoonright
A^+)\cup(G^{-1}\upharpoonright A-A^+)$. I claim that this
is the desired bijection between $A$ and $B$. First, it is
clearly a class that is definable from $F$ and $G$. Second,
it is a function from $A$ to $B$. Note that if $a\in A\_n$,
then $G(F(a))\in A\_{n+1}\subset A^+$, and so $F(a)$ is not
$G^{-1}(a')$ for any $a'\in A-A^+$. Thus, the function $H$
is injective. Secondly, if $b\in B$ and $G(b)\in A\_n$, then
it must be that $n\geq 1$ and so $b=F(a')$ for some $a'\in
A\_{n-1}\subset A^+$, putting $b\in\operatorname{ran}(H)$. Otherwise,
$G(b)\notin A^+$, and so again $b\in\operatorname{ran}(H)$. So $H$ is a
bijection. QED

Some of the other proofs can also be formalized for
classes, if one simply uses the sequences as I did here for
iterating.
To ask this question, you have to first be clear what you mean by a "class". Do you mean a finite formula in ZFC language with one free variable, P(x)? (Writing P(x) roughly means "x has property P".)

Second, what do you mean by a map from a class P to a class Q? Do you mean a class of ordered pairs?

If "yes" to both of the above, then I think the answer to your question is also "yes", because the Schroeder-Bernstein argument will allow you to explicitly write down a formula F(x,y) from the formulas P( ) and Q( ) that is a "bijection" in the sense that for all x such that P(x) there is a unique y such that Q(y) and F(x,y).

If, however, you mean class in some other sense, like the undefined notion of "class" used in NBG set theory, or something more vague, a different answer will be required.

I recommend reading the wikipedia article on [Zermelo-Frenkel Set Theory](http://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory)
 and browsing related articles until you feel comfortable the precise meanings of the basic terminology :)