0-tensor field is the real value function I was reading Lee's smooth manifold,in page 317.
there is a relation about the tensor bundle that is:
$$T^{(0,0)}TM = M \times \mathbb{R}   \tag{1}$$
then the book says:0-tensor field is the same as a continuous real value function.

The question are two :

*

*why (1) holds

*how to see the 0-tensor field on the tangent bundle as a real value function.


for the second question :By definition tensor field is section $A:M \to T^{(0,0)}TM$
But if (1) holds $A:M \to M\times \mathbb{R}$ is not a real value function as $M\to \mathbb{R}$?
(where bundle of mix tensor is defined as $T^{(k, l)} T M=\coprod_{p \in M} T^{(k, l)}\left(T_{p} M\right)$)
I don't know how to make this two question very clear.
 A: The first point holds because
$$T^{(0,0)}TM=\coprod_{p\in M}T^{(0,0)}(T_pM)=\coprod_{p\in M}\mathbb{R}=M\times\mathbb{R},$$
where the second equality is a convention ($T^{(0,0)}(T_pM)$ is $0$ copies of $T_pM$ and $0$ copies of $(T_pM)^*$, as usual you chose "$0$ copies" to mean the identity element for the operation you are dealing with - there it is $\mathbb{R}$ since the operation is the tensor product over $\mathbb{R}$).
The second point is that the map $A:M\to M\times\mathbb{R}$ has to check $\pi\circ A=\mathrm{id}_M$, where $\pi:M\times \mathbb{R}\to M$ is the first projection, since by definition a tensor field is a section of your bundle. So a $0$-tensor field is written $A(x)=(x,f(x))$, and so they are the same as real-valued functions if you use the identification
$$\varphi:\Gamma(T^{(0,0)}(TM))\to C^\infty(M,\mathbb{R}),\varphi(A)=\mathrm{pr}_2\circ A,$$
where $\Gamma(T^{(0,0)}(TM))$ is the set of $0$-tensors (i.e. the set of all sections of the bundle $T^{(0,0)}(TM)$) and where $\mathrm{pr}_2:M\times\mathbb{R}\to\mathbb{R}$ is the second projection.
