*This question is of course related to [this earlier MO question](https://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold), but I don't believe is answered by the posts there.*


My favorite proof of the Cantor-Schroeder-Bernstein theorem actually establishes something stronger: if $f:A\rightarrow B$ and $g:B\rightarrow A$ are injections, then there is a bijection $h:A\rightarrow B$ contained in the union $f\cup g^{-1}$. This is an extremely strong property. For example, let $A$ and $B$ each be copies of $\mathbb{N}$ as a linear order and consider the embeddings $f:A\rightarrow B:x\mapsto 2x$ and $g: B\rightarrow A: x\mapsto 3x$. Then $A\cong B$, but the unique isomorphism from $A$ to $B$ can't be pieced together from $f$ and $g$ in any reasonable way, let alone literally being $\subseteq f\cup g^{-1}$.


I'm interested in any situations where a strengthening of CSB along the above lines holds. To keep things reasonably narrow, the following potential CSB strengthening for vector spaces seems natural to consider but unlikely to hold:



> 
> Suppose $f:V\rightarrow W, g:W\rightarrow V$ are linear embeddings between disjoint vector spaces over a field $k$. Let $E$ be the equivalence relation on $V\sqcup W$ generated by (the graphs of) $f$ and $g$. For $a\in V$ (resp. $b\in W$) let $\widehat{a}=\langle W\cap [a]\_E\rangle\_W$ (resp. $\widehat{b}=\langle V\cap [b]\_E\rangle\_V$), where "$\langle \cdot\rangle\_U$" means "span in $U$."
> 
> 
> Must there be an isomoprhism $h:V\cong W$ such that for all $a\in V$ and all $b\in W$ we have $h(a)\in \widehat{a}$ and $h^{-1}(b)\in\widehat{b}$?
> 
> 
> 



The scope of a "strong Cantor-Bernstein" property
For infinite-dimensional vector spaces over any field, there is no such isomorphism. We will handle the countable-dimensional case, as the others are similar. Let $V$ be generated by the linearly independent vectors $v\_0,v\_1,\ldots$, and similarly let $W$ be generated by the linearly independent vectors $w\_0,w\_1,\ldots$. Let $f$ be the linear map uniquely determined by taking $v\_i\mapsto w\_{i+1}$, and similarly let $g$ be the linear map uniquely determined by taking $w\_i\mapsto v\_{i+1}$. (These might be termed "right-shift maps".) They are both injective.


Let $h\colon V\to W$ be any vector space isomorphism.


Now, $[v\_0]\_{E}=\{v\_0,w\_1,v\_2,w\_3,\ldots\}$ and $[v\_1]\_{E}=\{w\_0,v\_1,w\_2,v\_3,\ldots\}$. So, if $h$ is supposed to have the stated property, then $h(v\_0)\in {\rm Span}(w\_1,w\_3,\ldots)$, while $h^{-1}(w\_0)\in {\rm Span}(v\_1,v\_3,\ldots)$.


However, $[v\_0+h^{-1}(w\_0)]\_{E}$ does not have $w\_0$ in the support of any element. So, we cannot have $w\_0$ in the support of $$h(v\_0+h^{-1}(w\_0))=h(v\_0)+w\_0\in w\_0+{\rm Span}(w\_1,w\_3,\ldots),$$ which gives us the needed contradiction.


