 Solution

Because $F:\mathbf{Set}\to \mathbf{Set}$ preserves finite limits and colimits, $F(1)$ is a singleton and $F(\varnothing )=\varnothing$. For $x\in X$, let $\bar{x}:1\to X$ select $x$ and definewhere $\ast$ is the unique element of $F(1)$. This is natural in $X$. Any natural transformation $1_{\mathbf{Set}}\to F$ has the unique possible component at $1$, and naturality along every $\bar{x}$ forces the displayed value, so $\alpha$ is unique.

If $x\ne y$, the equalizer of $\bar{x},\bar{y}:1\rightrightarrows X$ is empty. Since $F$ preserves this equalizer, $F(\bar{x})(\ast )\ne F(\bar{y})(\ast )$. Hence every ${\alpha }_X$ is injective, so $\alpha$ is pointwise monic.

Finite-colimit preservation makes ${\alpha }_n:n\to F(n)$ bijective for every finite set. Use the stated characterization of $\mathbb{N}$ by the coproduct diagramand the coequalizer of $s,1_{\mathbb{N}}:\mathbb{N}\rightrightarrows \mathbb{N}$. Applying $F$ preserves both diagrams. Naturality and uniqueness in this characterization identify ${\alpha }_{\mathbb{N}}:\mathbb{N}\to F(\mathbb{N})$ as an isomorphism.

For a countable family $(A_n)$, let $p:{\coprod }_n{A}_n\to \mathbb{N}$ record the summand. Each squareis a pullback. Applying $F$ and using $F(1)\cong 1$ and $F(\mathbb{N})\cong \mathbb{N}$ shows that $F(A_n)$ is exactly the fiber of $F(p)$ over $n$. Those fibers partition $F({\coprod }_n{A}_n)$, so the canonical mapis bijective. Thus $F$ preserves countable coproducts.

Now choose $x\in F(A)\setminus {\alpha }_A(A)$ and defineIt is upward closed. Since $A=A^{\prime }\bigsqcup (A\setminus A^{\prime })$ and $F$ preserves binary coproducts, $x$ lies in exactly one of the two summands, so exactly one of $A^{\prime }$ and its complement lies in $U$. Pullback preservation gives closure under finite intersections.

For countable completeness, take $A_n\in U$ and put $B={\bigcap }_n{A}_n$. If $B\notin U$, then $A\setminus B\in U$. Partition $A$ into $B$ and the setswhich record the first failed membership. Since $F$ preserves countable coproducts, exactly one cell of this partition lies in $U$. It cannot be $B$, so some $C_n\in U$. But $C_n\subseteq A\setminus A_n$, forcing $A\setminus A_n\in U$, contrary to $A_n\in U$. Hence $B\in U$.

Finally no finite $K$ belongs to $U$. Indeed, $F(K)\cong K$ through ${\alpha }_K$, so if $x$ came from $F(K)$ then naturality would put $x$ in the image of ${\alpha }_A$. Thus $U$ is a countably complete ultrafilter and is nonprincipal.

Conversely, let $U$ be such an ultrafilter on $A$ and define the ultrapower endofunctor of setsIt preserves the terminal object and products: the mapis bijective because $U$ is closed under finite intersections. It preserves equalizers because an equality holding for an equivalence class holds on a $U$-large set, and the representative can be changed off that set to land in the equalizer. Hence it preserves all finite limits.

For $f:A\to {\coprod }_n{B}_n$, countable completeness implies that one index fiberbelongs to $U$; otherwise the countable intersection of all complementary fibers would be empty and belong to $U$. Thus every class $[f]$ lies in one and only one $F(B_n)$, proving preservation of countable coproducts.

The class of the identity map $A\to A$ is not represented by a constant map, since every equality set $\{a:a=a_0\}$ is a singleton and is not in $U$. Therefore ${\alpha }_A:A\to F(A)$ is not surjective. Since $\alpha$ is the unique natural transformation from the identity functor to $F$, a natural isomorphism $1_{\mathbf{Set}}\cong F$ would have to equal $\alpha$, which is impossible.