Let $(X,E)$, $(Y,F)$ be Borel equivalence relations. Assume $(X,E)\leq_{\text{B}}(Y,F)$.${}^{\dagger}$ Does it hold that: $F$ hyperfinite $\Longrightarrow$ $E$ hyperfinite?${}^{\ddagger}$ I do not see how to prove this in an obvious way at least, since the Borel reduction does not provide an injection.
${}^{\dagger}$ That is, there is a (not necessarily injective) Borel man $\varphi:X\rightarrow Y$, which reduces the equivalence relation on $X$ to that on $Y$ as follows
$$(\forall{x,x’\in X})\quad{x’~E~x\Longleftrightarrow \varphi(x’)~F~\varphi(x)}.$$
${}^{\ddagger}$ A Borel equivalence relation $(Z,R)$ is said to be hyperfinite, exactly when there is a nondecreasing $(R_{n})_{n\in\omega}$ of finite Borel equivalence relations on $Z$, so that $\bigcup_{n\in\omega}R_{n}=R$.
I think that the answer is no. The argument is as follows: Equality on a Polish space is a finite Borel equivalence relation; so it is enough to reduce some equivalence relation with big classes to equality in order to get a counterexample.
Readily, take $Y:=\{0,1\}$ (discrete), $F=\mathrm{id}_Y$, $X$ any space with a nontrivial clopen partition into two pieces, and $E$ the relation given by this partition. For example, $X=[0,1]\cup[2,3]$ with the relative topology of $\mathbb{R}$, and $E:=[0,1]^2\cup [2,3]^2$. Now the map that sends $[0,1]$ to $0$ and $[2,3]$ to $1$ is a continuous (a fortiori, Borel) reduction of $E$ to $F$.