The recursion theorem

The set $N\times A$ is closed, so the collection of closed subsets of $N\times A$ is nonempty; let $g:=\bigcap \{R:R\subseteq N\times A\text{ closed}\}$, which is itself closed, since if every closed $R$ contains $(0,a)$ then so does $g$, and if $(n,x)\in g$ then $(n,x)\in R$ for each closed $R$, whence $(\sigma (n),f(x))\in R$ for each such $R$, so $(\sigma (n),f(x))\in g$.

Let $D=\{n\in N:\mathrm{\exists }x\in A,(n,x)\in g\}$; then $0\in D$ since $(0,a)\in g$, and if $n\in D$ with $(n,x)\in g$ then $(\sigma (n),f(x))\in g$ by closure, so $\sigma (n)\in D$; by induction $D=N$, hence $g$ relates every $n$ to at least one $x$.

Suppose $(0,b)\in g$ with $b\mathrm{\ne }a$; then $g\setminus \{(0,b)\}$ still contains $(0,a)$ and is closed, since any generated pair $(\sigma (m),f(z))$ has $\sigma (m)\mathrm{\ne }0$ by P1 and so is never equal to $(0,b)$; this contradicts the minimality of $g$, so $a$ is the unique second coordinate paired with $0$.

Let $U=\{n\in N:\text{there is exactly one }x\text{ with }(n,x)\in g\}$; then $0\in U$ by step 2.2.

Assume $n\in U$ with unique value $x$, so $(n,x)\in g$ and $(\sigma (n),f(x))\in g$ by closure; if $(\sigma (n),c)\in g$ with $c\mathrm{\ne }f(x)$, then $g\setminus \{(\sigma (n),c)\}$ contains $(0,a)$ because $\sigma (n)\mathrm{\ne }0$, and is closed, because a generated pair $(\sigma (m),f(z))$ with $(m,z)\in g\setminus \{(\sigma (n),c)\}$ can equal $(\sigma (n),c)$ only if $\sigma (m)=\sigma (n)$ and $f(z)=c$, but $\sigma (m)=\sigma (n)$ forces $m=n$ by injectivity, then $z=x$ by uniqueness at $n$, giving $f(z)=f(x)\mathrm{\ne }c$, a contradiction; hence $g\setminus \{(\sigma (n),c)\}$ is closed and smaller than $g$, contradicting minimality, so $f(x)$ is the unique value at $\sigma (n)$ and $\sigma (n)\in U$.

By induction $U=N$, so together with the totality from step 2.1 the relation $g$ is a function $g:N\to A$; it satisfies $g(0)=a$, and closure with uniqueness gives $g(\sigma (n))=f(g(n))$ for all $n$.

If $h:N\to A$ also satisfies $h(0)=a$ and $h(\sigma (n))=f(h(n))$, then the set $\{n:g(n)=h(n)\}$ contains $0$ and is closed under $\sigma$, since $g(n)=h(n)$ implies $g(\sigma (n))=f(g(n))=f(h(n))=h(\sigma (n))$, so by induction it equals $N$, that is $h=g$.