Boolean lattice - Interval sublattice

Question

I cannot conclude with the demonstration of the following exercise, I have already verified that the map is a homomorphism, but I do not know how to calculate $f(L)$.

For a Boolean lattice $B$ and $a, b \in B$ such that $a \leqslant b$, show that the interval sublattice $\displaystyle [a,b] := \uparrow \negmedspace a \cap \downarrow \negmedspace b = \{ x \in B | a \leqslant x \leqslant b \}$ is a Boolean lattice. [Hint. First show that for any distributive lattice $L$ the map $f:L \rightarrow L$, given by $f(x) := (x \lor a) \land b$, is a homomorphism. Then calculate $f(L)$.]

Thanks in advance.

Answer 1

Given that $a \leq b$, by distributivity we get $$f(x) = (x \vee a) \wedge b = (x \wedge b) \vee a,$$ whence $f(L) \subseteq [a,b]$.

For $x \in [a,b]$, $$f(x)=(x\vee a) \wedge b = x \wedge b = x,$$ and thus, $f(L)=[a,b]$.

Since $L$ is Boolean, in particular for $x \in [a,b]$, $$f(x) \wedge f(x') = f(x \wedge x') = f(0) = a,$$ and likewise $f(x) \vee f(x') = b$.

Hence $[a,b]$ is Boolean.

Answer 2

One doesn't "calculate" $f(L)$, one makes a reasonable guess (eg $f(L) \stackrel{?}{=} [a,b]$ ) and then one tries to establish the guessed equality.

$f(L) \subseteq [a,b]$ :

Let $x \in L$. Clearly, $f(x) \leqslant b$. We show that $f(x)\geqslant a$ :
Since $a\leqslant a\vee x$ and $a\leqslant b$, $a$ is a lower bound of $\{a\vee x,\, b\}$. Hence, $a \leqslant \min \{a\vee x,\, b\} = f(x)$.

Can you show the other inclusion? If no, tell me where you're stuck.