How is the category of real numbers defined using the order relation $\le$?

The symbol $\cong$ denotes an isomorphism in the category of sets, which means it is simply a bijection between the two hom-sets. Because hom-sets are sets, their isomorphisms are bijective functions.

Conditions: The compared objects are hom-sets; The ambient category for the isomorphism is the category of sets

The numbers 2 and π are used as concrete examples to illustrate how the inclusion functor $I$ and the floor functor $F$ act on objects. The lecturer computes $I(2) = 2$ and $F(\pi) = \lfloor \pi \rfloor = 3$ to build intuition for the adjunction before deriving the general inequality equivalence.

Conditions: The example is set in the categories $\mathbb{Z}$ and $\mathbb{R}$.; $I$ is the inclusion functor $\mathbb{Z} \to \mathbb{R}$.; $F$ is the floor functor $\mathbb{R} \to \mathbb{Z}$.

This statement is the concrete order-theoretic consequence of the adjunction between the inclusion functor $I$ and the floor functor $F$. It means that for any integer $a$ and real number $b$, the inequality $a \le \lfloor b \rfloor$ holds exactly when $a \le b$ holds.

Conditions: $a$ must be an integer.; $b$ must be a real number.; The equivalence is stated as the consequence of the adjunction example.

In the example, $I$ is the inclusion functor from the category of integers $\mathbb{Z}$ to the category of real numbers $\mathbb{R}$, and $F$ is the floor-function functor from $\mathbb{R}$ to $\mathbb{Z}$. The lecturer states that $I$ is left adjoint to $F$.

Conditions: Defined between the two order-categories $\mathbb{Z}$ and $\mathbb{R}$.; $I$ is the inclusion functor $\mathbb{Z}\to\mathbb{R}$.; $F$ is the floor-function functor $\mathbb{R}\to\mathbb{Z}$.

We cannot speak of a natural isomorphism directly between $F$ and $G$ because natural isomorphisms compare functors with the same domain and codomain. Instead, the video reformulates the condition: $F$ and $G$ are adjoint if and only if the two curried hom-functors $\mathcal{B}(F(-),-)$ and $\mathcal{A}(-,G(-))$, both mapping from $\mathcal{A}^{op} \times \mathcal{B}$ to SET, are naturally isomorphic.

Conditions: $F:\mathcal{A}\to\mathcal{B}$ and $G:\mathcal{B}\to\mathcal{A}$; Both sides are regarded as functors out of $\mathcal{A}^{op}\times\mathcal{B}$; The natural isomorphism is between functors with the same domain and codomain