There is an Equivalence Set to Ord

Let $\mathbf{Set}$ be Set (Category). Let $\mathbf{Ord}$ be Ord (Category). Then $\mathbf{Set}\simeq \mathbf{Ord}$ are Eqiuvalence (Category Theory).

Proof

Let $F:\mathbf{Set}\to \mathbf{Ord}$.

$$\begin{array}{r}\text{Take }\alpha :\mathbf{Set}\to \mathbf{Ord}\\ \text{s.t. }\alpha X\cong X\\ \text{Let }{\varphi }_X:\alpha X\cong X\\ F(X)=\alpha X\\ F(f:X\to Y)={\varphi }_Y\circ f\circ {\varphi }_X^{-1}\end{array}$$

Such $\alpha$ is constructable since there is an ordinal for every cardinality.

Full Functor

Given $X,Y:\mathbf{Set}$, Given $f:\alpha X\to \alpha Y$, Let $g={\varphi }_Y^{-1}\circ f\circ {\varphi }_X^{-1}$ Then $F(g)=f$

Faithful Functor

Given $f,g:X\to Y$ Suppose $F(f)=F(g)$ Then ${\varphi }_Y^{-1}\circ F(f)\circ {\varphi }_X={\varphi }_Y^{-1}\circ F(g)\circ {\varphi }_X$ So ${\varphi }_Y^{-1}\circ {\varphi }_Y\circ f\circ {\varphi }_X^{-1}\circ {\varphi }_X={\varphi }_Y^{-1}\circ {\varphi }_Y\circ g\circ {\varphi }_X^{-1}\circ {\varphi }_X$ This $f=g$.

Essentially Surjective Functor

Given $\beta :\mathbf{Ord}$ Have $\beta :\mathbf{Set}$. $F(\beta )=\alpha \beta$ Then by definition of $\alpha$ , $\alpha \beta \cong \beta$.