Why can $T_p \times T^{*}_{p} \rightarrow \Bbb{R}$ be thought of as $T_p \rightarrow T_p$ or $T^{*}_{p} \rightarrow T^{*}_{p}$? My lecture notes about tensors say the following,

A tensor of type [1, 1] is a map from from $T_p \times T^{*}_{p} \rightarrow \Bbb{R}$. We can also think of this as a map from $T_p \rightarrow T_p$ or $T^{*}_{p} \rightarrow T^{*}_{p}$. There is a special such map, the identity; the corresponding tensor is called the
Kronecker delta.

I am unsure as to why we can think of the map $T_p \times T^{*}_{p} \rightarrow \Bbb{R}$ as  $T_p \rightarrow T_p$ or $T^{*}_{p} \rightarrow T^{*}_{p}$. Please could someone explain this to me?
 A: There is something very important the lecture notes leave out: those are not just "maps", but rather they are bilinear maps or linear maps.
What you are asking about is a property of all finite-dimensional vector spaces $V$ over all fields $K$. The following types of maps are easily transformed into each other:
(1) bilinear maps $V \times V^* \to K$,
(2) linear maps $V \to V$,
(3) linear maps $V^* \to V^*$.
For example, if $B \colon V \times V^* \to K$ is bilinear let's build a linear map $L_B \colon V \to V$ from it.  For each $v \in V$, the mapping
$B(v,\cdot) \colon V^* \to K$ is linear, so $B(v,\cdot) \in V^{**}$. By double duality there's a natural isomorphism $V^{**} \to V$: every element of $V^{**}$ has the form $\varphi \to \varphi(w)$ for a unique $w \in V$. So for each $v \in V$ there is a unique $w \in V$ such that $B(v,\varphi) = \varphi(w)$ for all $\varphi \in V^*$. This $w$ is determined by $B$ and $v$, so let's write it as $L_B(v)$. Thus to each bilinear $B \colon V \times V^* \to K$ and $v \in V$ we get a unique $L_B(v) \in V$ such that
$$
B(v,\varphi) = \varphi(L_B(v))
$$
for all $\varphi \in V^*$.  It's left to you to check that $L_B \colon V \to V$ is linear.  This is how maps $B$ of the form (1) can be turned into maps $L_B$ of the form (2).
The correspondence $B \mapsto L_B$ is a mapping ${\rm Bil}_K(V,V^*) \to {\rm Hom}_K(V,V)$ from a space of $K$-bilinear maps to a space of $K$-linear maps. Those two $K$-vector spaces of maps have the same finite dimension $(\dim_K V)^2$, so to show the maps of type (1) are in bijection with maps of type (2), show $B \mapsto L_B$ is a linear mapping ${\rm Bil}_K(V,V^*) \to {\rm Hom}_K(V,V)$ and then show it is either one-to-one or onto (either property implies the other by comparing the dimensions). It's much easier to show this is one-to-one (check the kernel is trivial).
By changing our focus to the second coordinate of $B$ instead of the first coordinate of $B$, and using the definition of the dual space of $V$ (each linear map $V \to K$ is a unique element of $V^*$), for each $\varphi \in V^*$ there is a unique $\psi \in V^*$ such that
$B(v,\varphi) = \psi(v)$ for all $v \in V$. This $\psi$ depends on $B$ and $\varphi$ so write it as $M_B(\varphi)$. That is,
$$
B(v,\varphi) = (M_B(\varphi))(v)
$$
for all $v \in V$. Check the bilinearity of $B$ implies $M_B \colon V^* \to V^*$ is linear. This is how maps of the form (1) can be turned into maps of the form (3), and by checking $B \mapsto M_B$ is a linear map
${\rm Bil}_K(V,V^*) \to {\rm Hom}_K(V^*,V^*)$ between two spaces of the same finite dimension that is one-to-one, it is a bijection.
A: Generally, a function $f \in C^{A\times B},$ i.e. $f : A \times B \to C$ can be identified with a function $A \to C^B$: if we give $f$ an argument $a\in A$ then $f(a,\bullet) : B \to C,$ i.e. $f(a,\bullet)\in C^B.$ Similarly we can also identify $f$ with a function $B \to C^A.$
Now we only have to limit this to linear maps and use some congruence relations between linear spaces.
So let $F$ be a field ($\mathbb{R}$ or $\mathbb{C}$), $V$ be a vector space and $V^*$ its dual. Then a bilinear function $f : V \times V^* \to F$ can be identified with a linear function $V \to L(V^*,F)$ or a linear function $V^* \to L(V,F),$ where $L(A,B)$ denotes the space of linear maps from $A$ to $B$. But in finite dimensions, $L(V,F)=V^*$ and $L(V^*,F)=(V^*)^*\cong V.$
