# basis of tangent space of a submanifold defined as a graph

Let $$f : \mathbb{R}^2 \rightarrow \mathbb{R}$$ be a smooth function.

Let $$G:=\{(x, y, f(x, y)) : x,y \in \mathbb{R}^2\}$$ be its graph.

Find a basis for $$T_pG$$ for a $$p(x,y,z) \in G$$.

What I did:

I am trying first to determin $$T_pG$$.

f is smooth so $$df: T\mathbb{R}^2 \cong \mathbb{R}^2 \rightarrow T\mathbb{R} \cong \mathbb{R}$$.

And since $$T_pG=Im(v \rightarrow (v,df_0(v))$$ for $$v\in T_pG$$, can we say that $$T_pG=Im(v \rightarrow (v,df_0(v))$$ for $$v\in\mathbb{R}^2$$?

and if yes, how can we go from there? I think I miss something fundamental in understanding manifiolds defined by graph.

Thank you!

** $$\textbf{EDIT}$$ **

Following up on Andeas's answer, let's consider the map $$g:(x,y) \rightarrow (x,y,f(x,y))$$

g is clearly injective, smooth with an inverse $$h:(x,y,z) \rightarrow (x,y)$$ that is smooth so g is an homeomorphism over its image $$g(\mathbb{R}^2)$$.

In addition, g is smooth and $$dg: T\mathbb{R}^2 \cong \mathbb{R}^2 \rightarrow TG$$. dg is clearly injective so g is an immersion.

We may conclude that g is a parametrization of G and that a basis of $$TG$$ is the image of $$\{(1,0),(0,1)\}$$ by g, i.e.:

$$\{(0,1,f(0,1)) (1,0,f(1,0))\}$$.

Hint: I think it will be easier to approach the problem by observing that you can nicely parametrize the graph by the map $$\mathbb R^2\to G$$ defined by $$(x,y)\mapsto (x,y,f(x,y))$$. Then you can look at the image of the standard basis of $$\mathbb R^2$$ under the derivative of this parametrization.
Edit (based on the edit of the question):The first part of what you are writing is fine.But then you should use $$dg$$ rather than $$g$$ in order to identify the tangent spaces of $$G$$. The resulting basis for the tangent space $$T_{(x,y,f(x,y))}G$$ is $$\{(1,0,\frac{\partial f}{\partial x}(x,y)), (0,1,\frac{\partial f}{\partial y}(x,y))\}$$. In a basis free way, you can describe the tangent space as $$\{(v,df(x,y)(v)):v\in\mathbb R^2\}$$. You can also view this as the fact that the tangent space to the graph of $$f$$ in a point $$(x,y)$$ is the graph of $$df(x,y)$$.