I am currently reading through a book on mathematical analysis and I don't understand why this is "given" without a proof, as I don't see the intuitive nature of it.
Define $M$ as a set, $P(M)$ as the power set of $M$, $id_M: M\to M$ the identity function of $M$, given by $\text{id}_M(x)=x$ for all $x \in M$.
Given any function $f:X\to Y$, define $$P(f) : P(X)\to P(Y), \ \ \ A\mapsto P(f)(A)$$ as $P(f)(A)=f(A)$ for all proper subset $A$ of $X$.
My question: Why is $P(\text{id}_X)=\text{id}_{P(X)}$?
I guess that $\text{id}_{P(x)} = x$ for all $x \in P(X)$? And that $P(\text{id} x) = \{A | A \subset \text{id}x\}$, $\text{id}P(x) = \{a |A \subset X\}$ (not sure about this one)?
Now I kind of need to figure out how to show that $\text{id}x = X$ maybe and then it magically explains it all?
We need to show that $P(\operatorname{id}_ M)=\operatorname{id} _{P(M)}$. To show that two functions are equal, it is enough to show that their domain is equal and Elements of the domain are mapped to the same image.
Clearly $\operatorname{dom}\operatorname{id} _{P(M)} = P(M)$ by definition of $\operatorname{id}_{P(M)}$. And by definition of $P(f)$ we have $\operatorname{dom}P(\operatorname{id}_M)=P(M)$ as well.
Now we just need to show that for $A\in P(M)$ holds $P(\operatorname{id}_M)(A)=(\operatorname{id} _{P(M)})(A)$. $(\operatorname{id}_{P(M)})(A)=A$ is clear from definition of $\operatorname{id}_{P(M)}$ again. Now:
$$P(\operatorname{id}_ M)(A)=(\operatorname{id}_ M)(A)=\{(\operatorname{id}_ M)(a):a\in A\}=\{a:a\in A\}=A$$
First we use the definition of $P(f)$, then the definition of $f(A)$, then the definition of $\operatorname{id}_M$ and the last one is evident. The author thinks it's trivial because you only need to use one definition after another and don't really have to put too much thought into it. Any questions left?