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 $D = \{ n \in N : \exists x \in A,\ (n,x) \in g \}$; 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 $b \ne a$; then $g \setminus \{(0,b)\}$ still contains $(0,a)$ and is closed, since any generated pair $(\sigma(m),f(z))$ has $\sigma(m) \ne 0$ 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 $c \ne f(x)$, then $g \setminus \{(\sigma(n),c)\}$ contains $(0,a)$ because $\sigma(n) \ne 0$, 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 $f(z) = f(x) \ne c$, 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 $h = g$.