Adjoint of forgetful functor I want to solve an exercise about adjoints (but I am not sure about the steps).
Let  $U:\textbf{CL}\rightarrow\textbf{Posets}$ be the forgetful functor. Here we have a functor from the complete lattices with joinpreserving maps to posets. Let $F$ be the functor the other way around defined by:
$$F(X,\leq)=\{U\subseteq X:\forall x, y\in X: y\leq x\in U \Rightarrow y\in U\}$$ 
My question is: Why is the image under $F$ already an complete lattice? Why is F the left adjoint of U?
Someone with hints of solutions?
Thanks :-)
 A: Probably the author of the question now knows the answer to his/her question, but might as well answer it for future reference. 
First of all, why is $F(X)$ a complete lattice ? If $S\subset F(X)$, then $\bigcup S \in F(X)$. Indeed, if $x\in \bigcup S, y\in X, y\leq x$, then there is some $Y\in S$ with $x\in Y$. Therefore, $y\in Y$ (as $Y\in F(X)$) and so $y\in \bigcup S$. It is then clear that $\bigcup S$ is the supremum of $S$ in $F(X)$. 
Since all suprema exist in $F(X)$, this implies that $(F(X), \subset)$ is a complete lattice, which answers your first question. Now why is it the case that $F\dashv U$ ? 
Let $(X,\leq)$ be a poset, and $(L,\leq)$ a complete lattice. Assume you have a monotone map $f:X\to UL$. This is just a monotone map $f:X\to L$. Then we can define $f^* : F(X) \to L$ by $f^*(Y) = \sup f[Y]$ (well defined as $L$ is complete). It remains to show that $f^*$ preserves joins (it is clearly monotone). 
So let $S\subset F(X)$, and let $Y= \bigcup S$.We want to show $f^*(Y) = \sup f^*[S]$. First of all, let $f^*(T) \in f^*[S]$. Then $T\subset Y$, so $f^*(T)\subset f^*(Y)$. 
Conversely, let $P\in L$ be an upper bound of $f^*[S]$. We want to show $P\geq f^*(Y)$, but remembering the definition of $f^*$ this is just to say that $P$ is an upper bound of $f[Y]$. Let $y\in Y$. Then there is some $T\in S$ with $y\in T$. But $P\geq f^*(T) \geq f(y)$ so $P\geq f(y)$, and this shows that $P$ is an upper bound of $f[Y]$, which concludes the proof. 
Now given a monotone $g: X\to UL$, if $g^* = f^*$ then in particular, for $x\in X$, denoting $x\downarrow = \{y\in X : y\leq x\}$ which is clearly in $F(X)$, we have $g^*(x\downarrow) = f^*(x\downarrow)$. But $f,g$ are monotone so these are just $g(x)$ and $f(x)$ so $g(x) = f(x)$ and so $g=f$. So $f\mapsto f^*$ is injective.
Finally, let $k: F(X) \to L$ be a sup-preserving map (in particular a monotone one). Define $k_{\sharp}: X\to UL$ by $k_{\sharp}(x) = k(x\downarrow)$. Using that $k$ is sup-preserving we deduce that $k=(k_\sharp)^*$. Indeed, if $Y\in F(X)$ then $Y= \bigcup\{x\downarrow : x\in Y\}$, and so $k(Y) = \sup \{k(x\downarrow), x\in Y\} = \sup\{k_\sharp(x), x\in Y\} = \sup k_\sharp[Y] = (k_\sharp)^*(Y)$. This proves that $f\mapsto f^*$ is surjective, and so is a bijection $\mathrm{Hom}_{\mathbf{Posets}}(X,UL)\to \mathrm{Hom}_{\mathbf{CL}}(F(X), L)$. Moreover this bijection is clearly natural in both arguments, and so we may conclude: $F\dashv U$
