Covariant Power Set Functor being injective on arrows.

Question

Consider the covariant power set functor $$\mathcal{P}(A\xrightarrow{\ f\ }B) = \mathcal{P}A\xrightarrow{\ \mathcal{P}f\ }\mathcal{P}B$$ where $\mathcal{P}A$ is the power set of $A$, and for $X\subseteq A$, $\mathcal{P}f(X)$ is defined as: $$\mathcal{P}f(X) = \{f(x)\ |\ x\in X \}.$$ Could someone explain why this is injective on morphisms? From my understanding, this means that for any two morphisms $$\mathcal{P}f,\mathcal{P}g,$$ if $\mathcal{P}f=\mathcal{P}g$, then $f=g$. Which would imply if $\mathcal{P}f$ and $\mathcal{P}g$ agree on all subsets of $A$, then $f=g$. Or, for all $X\subseteq A$ if $f(X)=g(X)$, then $f\equiv g$. However, I was told that the last sentence is not true in general, so there must be something I am missing. Thanks.

Answer 1

Assume $f \neq g$, i.e. there is $x \in A$ such that $f(x) \neq g(x)$. Then $$\mathcal{P}f(\{x\}) = \bigl\{f(x') | x'\in\{x\}\bigr\} = \{f(x)\}$$ and $$\mathcal{P}g(\{x\}) = \bigl\{g(x') | x'\in\{x\}\bigr\} = \{g(x)\}$$ but since $f(x)\neq g(x)$ these aren't the same, i.e. $\mathcal{P}f$ and $\mathcal{P}g$ disagree on $X = \{x\}$.