prove $F$ is Independent of the choice of representative, and find $|[h]_{R_{h}}|$ as $h\in\mathbb{N}^{\mathbb{N}}$ and $R$ is an Equivalence relation

Question

Given a function $h:\mathbb{\ N}\longrightarrow\mathbb{N}$. Let $R$ be an Equivalence relation over $\mathbb{N}^{A}$, as $A\neq\emptyset$.

$$R_{h}= \{ \left\langle f,g\right\rangle \in\left(\mathbb{N}^{A}\times\mathbb{N}^{A}\right)|\ h\circ f=h\circ g \}$$

Prove:

1. $ \{f\in\mathbb{N^{N}}\vert\exists\ c\in\mathbb{N}:\ \forall n\in\mathbb{N\ }f(n)=c\} = \{h\in\mathbb{N^{N}}\vert\ |\mathbb{N}^{A}/R_{h}|=1\}$ meaning - the set of function $h \in \mathbb{N^{N}}$ such that the cardinality of the set $\ \mathbb{N}^{A}/R_{h}$ equals to $1$.

for the next two sections suppose that $A=\mathbb{N}\ $ and $\ h:\mathbb{N}\longrightarrow\mathbb{N}$, defined by:

$$h(n)=\left\{ \begin{array}{cc} n\ \ \ \ \ \ \text{if } & n\in\mathbb{N_{\mathit{even}}}\\ n-1\ \ \ \ \ \text{if}\ & n\in\mathbb{N}_{odd} \end{array}\right.$$

$$F:\mathbb{N}^{\mathbb{N}}/R_{h}\longrightarrow\mathbb{N}^{\mathbb{N}},\ \ \ \left(F(\left[f\right]_{R_{h}})\right)(n)=\lfloor\frac{f(n)}{2}\rfloor$$

2. Prove that the definition of the function $F$ is independent on the choice of representatives.
3. Find: $$\vert[h]_{R_{h}}\vert$$

my attempt:

section (1) I solved by proving that $$\{f\in\mathbb{N^{N}}\vert\exists\ c\in\mathbb{N}:\ \forall n\in\mathbb{N\ }f(n)=c\} \subseteq \{h\in\mathbb{N^{N}}\vert\ |\mathbb{N}^{A}/R_{h}|=1\}$$ and $$\{h\in\mathbb{N^{N}}\vert\ |\mathbb{N}^{A}/R_{h}|=1\}\subseteq \{f\in\mathbb{N^{N}}\vert\exists\ c\in\mathbb{N}:\ \forall n\in\mathbb{N\ }f(n)=c\}$$

by words, I proved that every constant function $\in \mathbb{N}^{\mathbb {N}}$ has only one Equivalence set.

In section 2 and 3, I don't know from where to start... help will be appreciated.

Answer 1

1)

This comes to verification of:$$R_h\text{ has exactly one equivalence class }\iff h\text{ is a constant function}$$

The LHS is evidently equivalent with: $$\text{ for all functions }f,g:A\to\mathbb N\text{ we have }h\circ f=h\circ g\tag1$$

It is immediate the RHS implies $(1)$ and conversely since $A\neq\varnothing$ it can be shown that also $(1)$ implies the RHS. If $h$ is not constant then there are elements $n,m\in\mathbb N$ with $h(n)\neq h(m)$ and prescribing $f:A\to\mathbb N$ by $a\mapsto n$ and $g:A\to\mathbb N$ by $a\mapsto m$ we achieve that $h\circ g\neq h\circ g$.

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2)

First prescribe $k:\mathbb N\to\mathbb N$ by $n\mapsto\lfloor n/2\rfloor$ and prescribe $K:\mathbb N^{\mathbb N}\to\mathbb N^{\mathbb N}$ by $f\mapsto k\circ f$.

That means exactly that $K(f)(n)=\lfloor f(n)/2\rfloor$.

Now observe that $2k=h$ so that the functions $k$ and $h$ determine eachother and: $$fR_hg\iff h\circ f=h\circ g\iff k\circ f=k\circ g\iff K(f)=K(g)\tag2$$

This means that function $K$ respects relation $R_h$ hence can be written as $$K=F\circ\lfloor\;\rfloor$$ where $F:\mathbb N^{\mathbb N}/R_h\to\mathbb N^{\mathbb N}$ is well defined.

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3)

$(2)$ tells us that this comes to finding the cardinality of $$[h]_{R_h}=\{f\in\mathbb N^{\mathbb N}\mid K(f)=K(h)\}$$

So it concerns the functions $f:\mathbb N\to\mathbb N$ that satisfy $\lfloor f(n)/2\rfloor=\lfloor h(n)/2\rfloor$.

Note that $h$ only takes even values so that $\lfloor h(n)/2\rfloor=h(n)/2$ and you can focus on functions $f:\mathbb N\to\mathbb N$ that satisfy:$$2\lfloor f(n)/2\rfloor=h(n)$$

Give that a try yourself.