Let $X$ a Tychonoff space, $(e, \beta X)$ Stone–Čech compactification and consider $(i,K)$ another compactification of $X$ that satisfies:
$\forall f: X \to \mathbb R$ continuous and bounded, exists $\overline f : K \to \mathbb R$ continuous with $\overline f \circ i = f$.
Show that $(i,K)$ and $(e, \beta X)$ are equivalent compactifications.
My attempt: It's sufficient showing that for all Hausdorff compact $L$ and $\forall f: X \to L$ continuous, exists $g: K \to L$ continuous with $g \circ i = f$.
Let $L$ a Hausdorff compact and $g: X \to L$ a continuous function. (My problem is here) If I could consider $h: L \to \mathbb R$ continuous and injective, we'd have $f = h \circ g : X \to \mathbb R$ is continuous and bounded (because $h \circ g (X) \subset h(L)$ and $h(L)$ is compact in $\mathbb R$), by hypothesis, $\exists \overline f : K \to \mathbb R$ continuous with $\overline f \circ i = f$. Then $\overline f \circ i = h \circ g$, so if I'd have $h^{-1}: h(L) \to L$ continuous, $(h^{-1} \circ \overline f )\circ i = g$.
I'm not sure I can consider a function $h: L \to \mathbb R$ and why $h^{-1}$ would be continuous.