Let $f:V\rightarrow M$ be a smooth map between smooth manifolds.
Why is $Gr(f):=\{\ \left(v,f(v)\right)\ |\ v\in V\} $ a submanifold of $V \times M$ ?
This is used in a proof by Kosinski ("Differentiable Manifolds", chapter 4, Corollary 2.5) but it is not justified.
Let $A:=\{(x,x)~\vert x\in M\}$. From the definition of the smooth structure on the product $M\times M$, it's pretty clear that $A$ is a smooth submanifold of $M\times M$. Let $\Phi:V\times M\to M\times M$ be the smooth map defined by $\Phi(x,y)=(f(x),y)$. We have that
$$Gr(f)=\Phi^{-1}(A).$$
With the natural identification of $T_{(x,y)}(V\times M)=T_xV\times T_yM$, you have that for any $(h,k)\in T_{(x,y)}(V\times M)$,
$$d_{(x,y)}\Phi(h,k)=(d_xf(h),k).$$
Now the tangent space of $A$ is simply $T_{(x,x)}A=\{(h,h)\vert h\in T_x M\}$. Therefore for any $(x,y)\in V\times M$ you have that
$$T_{\phi(x,y)}A +d_{(x,y)}\Phi \left(T_{(x,y)}(V\times M)\right)=T_{\Phi(x,y)}(M\times M)$$ which just mean that $\Phi$ is transverse to the submanifold $A$. The last equality holds because $\{(0,k)\vert k\in T_y M\}\subset d_{(x,y)}\Phi \left(T_{(x,y)}(V\times M)\right)$.
You can conclude by using the following theorem, which can be found in chapter $1$ of M.W.Hirsh Differential Topology for example.
Theorem: Let $f:M\to N$ be a smooth map and let $A\subset N$ be a smooth submanifold. If $f$ is transverse to $A$ then $f^{-1}(A)$ is a smooth submanifold of $M$.
You can also take a look at this post which is closely related.
