# Some questions about set of bijections from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$

Let $$S(\mathbb{N}\times\mathbb{N}, \mathbb{N})$$ the set of bijective functions from $$\mathbb{N}\times\mathbb{N}$$ to $$\mathbb{N}$$. I have some questions about this set.

Note that $$S(\mathbb{N}\times\mathbb{N}, \mathbb{N}) \subseteq \mathbb{N}^{\mathbb{N}\times\mathbb{N}}$$. Consider the discrete topology in $$\mathbb{N}$$, and $$\mathbb{N}^{\mathbb{N}\times\mathbb{N}}$$ with product topology.

• what are the autohomomorphisms of $$S(\mathbb{N}\times\mathbb{N}, \mathbb{N})$$.
• It is possible to define an operation on $$S(\mathbb{N}\times\mathbb{N}, \mathbb{N})$$, such that it is a semigroup (or group).

The reason for my questions is that we know several things about $$S_\infty(\mathbb{N})$$ (set of bijective functions from $$\mathbb{N}$$ to $$\mathbb{N}$$), I was wondering if it is possible to transfer some results from $$S_\infty(\mathbb{N})$$, to $$S(\mathbb{N}\times\mathbb{N}, \mathbb{N})$$. At least topologically $$S_\infty(\mathbb{N})$$ and $$S(\mathbb{N}\times\mathbb{N}, \mathbb{N})$$ are the same, but I can't see what the autohomeomorphisms of $$S(\mathbb{N}\times\mathbb{N}, \mathbb{N})$$ look like, for example.

• There's no reason to consider $\mathbb{N}\times\mathbb{N}$, since $S(\mathbb{N}\times \mathbb{N}, \mathbb{N})\cong S(\mathbb{N}, \mathbb{N}) = S_\infty(\mathbb{N})$ using the bijection of $\mathbb{N}\times\mathbb{N}$ with $\mathbb{N}$. This also means that the set of autohomeomorphisms will be the same, up to application of this homeomorphism. Nov 20 at 20:57