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

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> 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?
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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: 
Ignoring set-theoretic technicalities of formulating the question properly, I see no reason that the usual proof of Schroder-Bernstein wouldn't work.

(Set-theoretic technicalities: In the standard language of set theory, you can't quantify over classes, so you can't quite state this. However, you can prove a metatheorem saying that whenever you exhibit two such injections, you can prove there is also a bijection. Alternatively, you could work in [set theory with classes](http://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory), in which the statement can be made properly and you ought to be able to prove it just like ordinary Schroder-Bernstein. Alternatively, it is a trivial corollary of the "global" axiom of choice (which implies, in particular, that all proper classes have the same size), though this is kind of applying a sledgehammer.)
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:

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> 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 $x \in A$
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> $G(x)=F(x,G|\_{\{y \in A : y R x\}})$.
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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)?