Could anyone check my working please?
A functor must fulfill the following criteria.
- To each $\mathscr C$-object $a$, a $\mathscr D$-object $F(a)$ and
- To each $\mathscr C$-arrow $f:a\to b$ a $\mathscr D$-arrow $F(f):F(a)\to F(b)$ such that
- $F(1_a)=1_{F(a)}$ for all $\mathscr C$-objects $a$
- $F(g\circ f)=F(g)\circ F(f)$, whenever $g\circ f$ is defined
The identity functor $1_{\mathscr C}:\mathscr C \to \mathscr C$ that has $1_{\mathscr C}(a)=a, 1_{\mathscr C}(f)=f$.
For 1, from left to right: $1_{\mathscr C}(1_a)=1_{\mathscr C}(a)=a=1_{\mathscr C}(a)=1_{1_{\mathscr C}(a)}$. The case for an arbitrary function $f$ is similar.
For 2(a), suppose we have $1_{\mathscr C}(1_a)$. But then $1_a$ just takes us back to $a$, then $1_{\mathscr C}$ by definition maps it to $a$. If we start off with $1_{1_{\mathscr C}(a)}$,$1_{\mathscr C (a)}=a$, and $1_a=a$ too. So regardless of which side we begin with, we still end up with $a$.
Now for 2(b) suppose we have $1_{\mathscr C}(g\circ f)$, but by definition that gives us $g\circ f$. On the other hand if we have $1_{\mathscr C}(g)\circ 1_{\mathscr C}(f)$, since $1_{\mathscr C}(g)=g$ and like wise for $f$, we also end up with $g\circ f$.
Power set functor: $\mathscr P: Set \to Set$ maps each set $A$ to its powerset $\mathscr P(A)$, and each function $f:A\to B$ to the function $\mathscr P (f): \mathscr P (A)\to \mathscr P (B)$ from $\mathscr P(A)$ to $\mathscr P(B)$ that assigns to each $X\subseteq A$ its $f$-image $f(X)\subseteq B$.
For 1, if the $\mathscr C$ object is an arbitrary set $A$ in the category $Set$, then the functor $\mathscr P$ by definition assigns $A$ to $\mathscr P(A)$. Likewise if the object is a function $f$, it is also well defined according to the definition of functor as per its definition.
For 2(a), suppose we have $\mathscr P(1_a)$. Since we are working with the category $Set$, $a$ would be a set $A$.$1_A$ would take us back to $A$, so $\mathscr P(1_a)=\mathscr P(A)$. On other hand, if we start from the RHS, $1_{\mathscr P(A)}=\mathscr P(A)$, so both LHS and RHS are identical to the same thing, ie. $\mathscr P(A)$.
For 2(b), suppose we have $\mathscr P(g\circ f)$. Let $f:A\to B$ and $g:B\to C$. Then $g\circ f: A\to C$, with $\mathscr P(g\circ f):\mathscr P(A) \to \mathscr P(C)$. i.e. $\mathscr P(g\circ f)$ assigns to each $X\subseteq A$ its $g\circ f$ image $g(f(X))\subseteq C$.
On the other hand, $\mathscr P(f): \mathscr P(A)\to \mathscr P(B)$ and $\mathscr P(g):\mathscr P(B)\to \mathscr P(C)$, therefore $\mathscr P(g)\circ \mathscr P(f): \mathscr P(A)\to \mathscr P(C)$.
$\mathscr P(f)$ assigns each $X\subseteq A$ to $f(X)\subseteq B$, and $\mathscr P(g)$ assigns each $Y\subseteq B$ to $g(Y)\subseteq C$. But because $f$ is a function from $A$ to $B$, if we take $f(X)=Y$, then $\mathscr P(g)\circ\mathscr P(f)$ assigns each $X\subseteq A$ to $g(f(X))\subseteq C$. Therefore given an arbitrary $X\subseteq A$ both $\mathscr P(g\circ f)$ and $\mathscr P(g)\circ\mathscr P(f)$ would output the same thing, i.e. $g(f(X))\subseteq C$.
Best Answer
For 2.a) it's about the $\mathscr P$-image of the identity function being itself an identity function.
By definition, we have $\mathscr P(1_A)(X)=1_A(X)=X$ for every $X\in \mathscr P(A)$, so it's indeed the identity function.
For 2.b) you should prove $\mathscr P(g\circ f) =\mathscr P(g)\circ \mathscr P(f)$, using again the definition of $\mathscr P$ on arrows.