How to prove that the map $f_a(x)=(x\wedge a, x\vee a)$ is onto

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I need to prove that, if $L$ is a distributive lattice and possesses $0$ and $1$, then $f_a:L\to \downarrow a\times \uparrow a$ defined by

$f_a(x)=(x\wedge a, x\vee a)$

is an isomorphism if and only if $a$ has a complement in $L$. (Note that $a\in L$).

I have already shown that if $f_a$ is an isomorphism, then the complement of a belongs to L.

Moreover, it's (not so) easy to prove that if a has a complement in L, then $f_a$ is one-to-one map.

Now I would like to demonstrate that $f_a$ is a surjection, i.e .: $\forall (\alpha,\beta)\in \downarrow a\times \uparrow a,\exists x\in L$ s.t. $f_a(x)=(\alpha,\beta)$, but I do not know how I can do it.

Any help is welcome.

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Hint: How would you solve $x\cap a=\alpha, x\cup a=\beta$ if $L$ was the Boolean algebra of subsets of a set $S?$