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This is something I never understood.

Let $A$ denote a set and let $2$ denote the 2-element set $\lbrace 0,1\rbrace$. Consider two (parallel) functions $f,g:A\to 2$. They correspond to subsets of $A$, and hence they form elements of the Boolean algebra $\mathcal{P}(A)$. Now $\mathcal{P}(A)$ can be viewed as a category: for any two objects $X,Y\in\mathcal{P}(A)$, there exists a unique arrow $X\to Y$ whenever $X$ is a subset of $Y$.

My question is: Just by having the set $A$ and the two functions $f$ and $g$, what needs to be satisfied by them, so that we get an arrow in the category $\mathcal{P}(A)$ between the objects (subsets) corresponding to $f$ and $g$?

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    @pepa.dvorak I would teach you what category theory is, but this comment box is too narrow :) – 2017-02-14
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    This definition seems unnecessarily abstract to me.. Instead of functions think of mere subsets and the thing gets easier. The "abstract nonsense" machinery will work the same. – 2017-02-14
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    In the first paragraph arrows are equivalent to "being a subset", so let's look at what "being a subset" corresponds to in the language of functions (let's maintain the classical definition "$a\in S \leftrightarrow f(a) = 1$" for $S \subseteq A$. If $g$ is another such function (representing a subset $T$), then $S \subseteq T$ if for any $a\in A$ such that $f(a)=1$, $g(a) =1$, too. And then there is an arrow $f\rightarrow g$ (Which is the same as in the answer of @MeesdeVries) – 2017-02-14

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We need $f \leq g$, that is, $$ \forall a \in A(f(a) \leq g(a)). $$ In other words, for each element $a \in A$, we need that if $a$ is in the subset defined by $f$, then it is also in the subset defined by $g$.

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    What does $\leq$ mean here? Is it inequality of natural numbers, viewed as elements of $2$, or is it inclusion of the image of $f$ to the image of $g$? – 2017-02-14
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    It means inequality of numbers. It corresponds to the subset relation on the side of the power set. – 2017-02-14