Let $X$ be a smooth manifold. Then as vector spaces $$C^{\infty} (X) \otimes C^{\infty} (X) \not\simeq C^{\infty} (X \times X).$$
I know that there is a map from the algebraic tensor product $C^{\infty} (X) \otimes C^{\infty} (X) \longrightarrow C^{\infty} (X \times X)$ defined on elementary tensors by $f \otimes g \mapsto (h : (x,y) \mapsto f(x) g(y))$ and then extend it to the whole of the algebraic tensor product $C^{\infty} (X) \otimes C^{\infty} (X)$ by imposing linearity. But I can't conclude injectivity and surjectivity of this map. Could anyone please help me in this regard?
Thanks for your time.
$\def\RR{\mathbb{R}}$Let $M$ be your $X$.
If $f:M\times M\to\RR$ is a function, for each $y\in M$ we may consider the function $f_y:x\in M\mapsto f(x,y)\in\RR$, which is a smooth function, and the subspace $V_f$ of $C^\infty(M)$ generated by $\{f_y:y\in M\}$.
If $f$ is in the image of the canonical map $C^\infty(M)\otimes C^\infty(M)\to C^\infty$, then $V_f$ is finite-dimensional.
To prove that your map is not surjective, then, it is enough to find a smooth function $f:M\times M\to\RR$ such that $V_f$ is not finite dimensional.
We may just as well look for such function with support contained in an open set of the form $U\times U$ with $U$ domain of a coordinate patch of $M$, and this reduces the problem to finding a function $f$ of compact support on $\RR^n\times\RR^n$ whose $V_f$ is not finite dimensional, and one can do this with the same idea as for $\sin(xy)$, times a bump function.
On the other hand, the map is injective. Suppose that $u=\sum_{i=1}^nf_i\otimes g_i$ is mapped to zero. We may suppose that the $n$ functions $f_1,\dots,f_n$ are linearly independent. We have that $\sum_if_i(x)g_i(y)=0$ for all choices of $x$ and $y$ in $M$. If $u$ is not zero, then there is a $j$ and a $y_0$ such that $g_j(y_0)\neq0$. We have that $\sum_if_i(x)g_i(y_0)=0$ for all $x$, and this is a non-trivial linear relation among the $f_i$, which is absurd.