dim as a functor

Let $\mathbb{F}-{\mathrm{V}}_{\mathrm{B}}$$\mathbb F-\mathrm{V_B}$ be the category of finite dimension vector spaces with basis.

Let $\mathbb{N}$$\mathbb N$ be a category. The object is $n=\{1,2,...,n\}$$n=\{1,2,...,n\}$. The morphism are functions between them.

Then $\dim$$\dim$ is a functor.

For the object, $(V,B)$$(V,B)$, the $\dim$$\dim$ map it to $(V,\{b_1,b_2,...,b_n\})\to n$$(V,\{b_1,b_2,...,b_n\})\to n$

For the morphism, $T$$T$ is determined at the basis. Hence $\dim (T)(b_i)=T(b_i)$$\mathrm{dim}(T)(b_i)=T(b_i)$ gives you the function between $n$$n$ and $m$$m$.

Hence $\dim$$\dim$ is functor.

 

 

Differentail neighbourhood and Taylor series

Consider $T_0:C^{\mathrm{\infty }}([-1,1])\to \mathbb{R}[[X]]$$T_0:C^{\infty}([-1,1])\to\mathbb R[[X]]$

$$\begin{array}{}\text{(1)}& T_0(f)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(0)}{n!}{x}^n\end{array}$$

The domain is not an integral domain, hence the kernel of $T_0$$T_0$ is a prime ideal, i.e. $\begin{array}{c}\bigcap _{n=1}^{\mathrm{\infty }}{m}_0^n\end{array}$$\bigcap_{n=1}^{\infty}m_0^n\\$.

Then $V\left(\bigcap _{n=1}^{\mathrm{\infty }}{m}_0^n\right)\cong \mathrm{Spec}(\mathrm{Im}(T_0))$$V\left(\bigcap_{n=1}^\infty m_0^n\right)\cong\mathrm{\mathrm{Spec}}(\mathrm{Im}(T_0))$. The only two ideals in $\mathrm{Im}(T_0)$$\mathrm{Im}(T_0)$ is $(0),m_0$$(0),m_0$.

The topology here is really simple, $m_0$$m_0$ is a closed set. This is the differential neighbourhood.