larger coherent family of functions One way to construct an Aronszajn tree is to build a sequence of functions $\langle e_\alpha : \alpha < \omega_1 \rangle$ such that $e_\alpha$ is an injection from $\alpha$ to $\omega$, and for any $(\alpha, \beta)$, $e_\alpha$ agrees with $e_\beta$ except on a finite set.
Is the following generalization possible?  There is a family of functions {$e_x : x \in [\omega_2]^\omega$} such that each $e_x$ is an injection from $x$ to $\omega$, and for any $x,y \in [\omega_2]^\omega$, {$\alpha : e_x(\alpha) \not= e_y(\alpha)$} is finite.
 A: To add to Joel's answer; you can construct such a family, so long as you're willing to relax the injectivity requirement to just "finite-to-one".  See P. Koszmider, "On Coherent Families of Finite-to-One Functions", JSL 58 (1993) no. 1, 128-138.  It's in JSTOR.
Koszmider proves that you can construct such a family (with functions that are finite-to-one) on any $[\omega_n]^\omega$ in ZFC, and any $[\kappa]^\omega$ ($\kappa$ an infinite cardinal) assuming $V = L$.
A: If the continuum hypothesis holds, then this is impossible. To see this, consider a family of $\omega_2$ many pairwise disjoint subsets of $\omega_2$, and for each $x$ in this family, consider the range of $e_x$, which is a subset of $\omega$. By the continuum hypothesis, there are only $\omega_1$ many possible ranges, and so there must be two disjoint countable subsets $x,y\subset\omega_2$ with $\text{ran}(e_x)=\text{ran}(e_y)$. Let $z=x\cup y$ and consider $e_z$. It cannot be that $e_z$ has only finite difference from $e_x$ and also from $e_y$, if $e_z$ is injective.
I'm not yet sure what happens when CH fails... 
Meanwhile, here are a few slightly weaker generalizations that are possible. 


*

*There is a family of injective functions $e_x:x\to \omega$ for all countable $x\subset\omega_1$, that is, for $x\in [\omega_1]^\omega$, such that $e_x$ and $e_y$ have at most finitely many disagreements on $x\cap y$. To see this, first fix a family of functions $e_\alpha$ for $\alpha\lt\omega_1$ as mentioned at the outset of the question, so that $e_\alpha$ injects $\alpha$ to $\omega$, and any two of them disagree only finitely. The point is that we may extend this family to $e_x:x\to \omega$ for $x\in[\omega_1]^\omega$ by defining $e_x=e_\alpha\upharpoonright x$, where $\alpha=\sup(x)$. The point is that if $\sup(y)=\beta$, then $e_\beta$ and $e_\alpha$ have only finitely many disagreements on their common domain, and so also $e_x$ and $e_y$ will therefore also have only finitely many disagreements. 

*There is a family of injective functions $f_x:x\to \omega$ for all $x\in(\omega_2)^\omega$, meaning $x\subset\omega_2$ and $x$ has order type $\omega$, such that $f_x$ and $f_y$ have at most finitely many disagreements on $x\cap y$. To see this, fix the $e_x$ as in the previous paragraph, and also for each $\beta\lt\omega_2$, fix an injection $\pi_\beta:\beta\to \omega_1$, and now for $x\subset\omega_2$ of order type $\omega$, let $\beta=\sup(x)$, and consider $\pi_\beta[x]\subset \omega_1$. We may let $f_x=(e_{\pi_\beta[x]}\circ\pi_\beta)\upharpoonright x$. That is, we use $\sup(x)$ to tell us how to regard $x$ as really a subset of $\omega_1$, and then we apply the corresponding function from before. The point is that if $x$ and $y$ are two increasing $\omega$ sequences in $\omega_2$, then if they have different suprema, then $x\cap y$ is finite anyway and we needn't worry about it, and if they have the same suprema, then $f_x$ and $f_y$ agree completely. 
