Is this map an embedding?

Question

can someone give me an idea why the map $F:M\rightarrow M\times \mathbb{R}$ s.t. $F(p)=(p,f(p))$ where $f:M\rightarrow \mathbb{R}$ is smooth, is an embedding?, a map s. t. $dF_p$ is injective and $F$ is homeo onto his image. I can not see anything using the differencial of $F$.

Answer 1

1. $F$ is smooth, since it is the (cartesian) product of smooth maps
2. $dF_p$ is injective, since $$dF_p=\begin{pmatrix}I_n&*\end{pmatrix}^t$$ so $rk(dF)=n$, and $dF$ is injective. This implies that $F$ is a smooth immersion
3. $F$ is injective (trivially, since $(p,f(p))=(q,f(q))\leftrightarrow p=q$).
4. $F: M\to F(M)$ is open, since for any open $A\subset M$, $F(A)=\left(A\times \mathbb{R}\right)\cap F(M)$.

Since every bijective open map is an homeomorphis, we have that $F$ is an embedding.