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: 
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 :)
This is an interesting question. I think there are some issues which the others did not mention yet. But I'm not an expert at all, I might be wrong. Please leave me a comment!

In the following, I work in $ZF$. Thus, a class is just a formula. There are some constructions and relations of sets which directly carry over to classes. For example $A=B$ means that the formulas of $A,B$ are equivalent.

Let's sketch the proof for classes. Let $A,B$ classes, $f : A \to B$, $g : B \to A$ injective maps. Define the classes $A\_n \subseteq A, B\_n \subseteq B$ recursively by $A\_0=A, B\_0 = B, A\_{n+1} = g[B\_n], B\_{n+1}=f[A\_n]$. Then $h : A \to B$, defined by $f$ on $\cap\_n A\_n \cup (\cup\_n A\_{2n} \setminus A\_{2n+1})$ and by $g^{-1}$ in the rest, is well-defined and a bijection.

Unions, cuts, images of functions etc. are not the problem. But what about the recursion? What we really need here is a recursion scheme for classes. Actually there is a theorem which might be called the transfinite recursion scheme for classes:

> 
> Let $R$ be a well-founded and set-like relation of the class $A$ and $F : A \times V \to V$ a function. Then there is a function $G : A \to V$, such that for all $x \in A$
> 
> 
> $G(x)=F(x,G|\_{\{y \in A : y R x\}})$.
> 
> 
> 

However, note that the images of $G$ are sets, not proper classes. We can't use that theorem here.

I think we need a meta-theorem stating that the above also holds when $V$ is replaced by the set of formulas and $R=\mathbb{N}$. Also, the meta-world should be able to talk about functions. But this is not plausible at all, since the resulting formula won't have to be finite, right?

For example, try to define a formula $G(n)$ recursively by $G(0)=\phi\_0$ (doesn't matter what $\phi\_n$ is) and $G(n)=G(n-1) \wedge \phi\_n$. Why should there be a formula $\psi(n,x)$ such that $\psi(n,x) \Leftrightarrow \wedge\_{i=0}^{n} \phi\_i$? I think we need a rather mighty logical calculus for that.

Also note that Francois' great answer [here](https://mathoverflow.net/questions/18042/axiom-of-infinity-needed-in-cantor-bernstein) (proving Schröder-Bernstein without the existence of the set $\omega$) also causes problems when you want to write down the formula for the part "...$\exists s : \{0,...,n\} \to A$ ...". Perhaps there is really no bijection between the two classes mentioned above (class of all singletons, and class of all 2-element sets)?