Injective function between $\kappa^{\omega}$ and $[\kappa]^{\leqslant \omega}$ Is there an injective function $\varphi :\kappa^{\omega} \rightarrow [\kappa]^{\leqslant \omega}=\{ A\subset \kappa :|A|\leqslant \omega\}$ such that $\varphi (\alpha) \backslash \varphi (\beta)$ and $\varphi (\beta)\backslash \varphi (\alpha)$ are both non-empty whenever $\alpha, \beta\in \kappa^{\omega}$, $\alpha \neq \beta$?
What if $\kappa =\aleph_{\omega}$ and GCH holds?
I'm trying to wrap my mind around a problem for my thesis and I really appreciate any help you can provide.
 A: (I’m assuming that $\kappa\ge\omega$.) Let $h:\omega\times\kappa\to\kappa$ be a bijection. Each $f\in{^\omega\kappa}$ is a subset of $\omega\times\kappa$, so let 
$$\varphi(f)=h[f]=\{h(\langle n,\xi\rangle):\langle n,\xi\rangle\in f\}\in[\kappa]^{\le\omega}\;.$$
Clearly $f\setminus g\ne\varnothing\ne g\setminus f$ whenever $f$ and $g$ are distinct members of ${^\omega\kappa}$, so $\varphi$ has the desired properties.
A: Here is a quick justification for $\kappa=\aleph_\omega$.
Recall that $\aleph_\omega^{\aleph_0}=\prod_{n<\omega}\aleph_n$. Therefore the product of the intervals $I_n=\omega_n\setminus\omega_{n-1}$ (take $\omega_{-1}=0$ for a uniform definition), satisfies that $\prod_{n<\omega}I_n$ has the same cardinality as $\kappa^\omega$.
However, every two distinct functions define distinct sets which are not $\subseteq$-comparable, since we choose one element from each interval and the intervals are pairwise disjoint.
A: The following is a variant of Brian's answer and I assume that $\kappa$ is a cardinal (well, I assume that $f^{*} \subseteq \kappa$ in the construction below. This is always the case if $\kappa$ is a cardinal, but we need way less.)
Let $\langle \ , \ \rangle \colon \operatorname{Ord} \times
\operatorname{Ord} \to \operatorname{Ord}$ be Gödel's pairing
function. For each $f \in \kappa^{\omega}$ let
$$
f^{*} := \{ \langle n, f(n) \rangle \mid n < \omega \} \in [\kappa]^{\omega}.
$$
The function $\pi \colon \kappa^{\omega} \to [\kappa]^{\leq \omega}, f
\mapsto f^{*}$ is clearly an injection and given distinct $f,g \in
\kappa^{\omega}$, we have $\langle n, f(n) \rangle \in f^{*} \setminus
g^{*}$ and $\langle n,g(n) \rangle \in g^{*} \setminus f^{*}$ for
any $n < \omega$ such that $f(n) \neq g(n)$. In fact, letting
$f \perp g := \{n < \omega \mid f(n) \neq g(n) \}$, we have
$$
f^{*} \setminus g^{*} = \{ \langle n, f(n) \rangle \mid n \in f \perp
g \}
$$
and
$$
g^{*} \setminus f^{*} = \{ \langle n, g(n) \rangle \mid n \in f \perp
g \}.
$$
