Semi-lattice and category theory

Question

How would one prove that a monotone map between two Semi-lattices that preserves the supernum for any subset implies the the map has a right adjoint and vice versa?

Answer 1

Given posets $A,B$, such that $B$ is complete, and a monotone map $g : B \to A$ we can prove:

$$g\text{ preserves infima} \Leftrightarrow g\text{ has a left-adjoint}$$

"$\Rightarrow$":

Define $f : A\to B$ by:

$$f(x) := \bigwedge \{ y\in B : x \leq g(y)\}\,\,\,(\bigwedge =\inf)$$

If $x\leq x'\in A$, then $\{y\in B : x\leq g(y)\} \supseteq \{y\in B : x'\leq g(y)\}$ hence, by applying $\bigwedge$, we have $f(x) \leq f(x')$. So $f$ is monotone.

If $x\leq g(y)$, then $y\in \{y'\in B : x\leq g(y')\}$ so $f(x) \leq y$. Conversely, if $f(x)\leq y$ then, by applying $g$ and using the fact, that $g$ preserves infima, we get $x\leq \bigwedge \{g(y')\in B : x\leq g(y')\} \leq g(y)$.

"$\Leftarrow$":

Let $(y_i)_{i\in I}$ be a family in $B$. Then $g(\bigwedge_{i\in I} y_j) \leq \bigwedge_{i \in I} g(y_i)$ just because $g$ is monotone. For the other inequality, we have $\bigwedge_{i\in I} g(y_i) \leq g(y_i)$ hence $f\left(\bigwedge_{i\in I} g(y_i)\right) \leq y_i$ for all $i\in I$. Therefore $f\left(\bigwedge_{i\in I} g(y_i)\right) \leq \bigwedge_{i\in I} y_i$, so $\bigwedge_{i\in I} g(y_i) \leq g(\bigwedge_{i\in I} y_i)$.

Using duality you get the statement for suprema and right adjoints.