When are two level sets of a function diffeomorphic? Let $f(x_1,\dots,x_n)$ be a polynomial over $n$ variables. Let $m(x_1,\dots,x_n)$ be a monomial in the same variables.
Are the surfaces $\{f=x_{n+1}\}$ and $\{f+m=x_{n+1}\}$ diffeomorphic?
My thought is that since we can easily deform one to the other by defining $f_s=f+s\, m$ with $s\in[0,1]$ and since there is not singularity to appear they have to be diffeomorphic. Is this true? If not, what can go wrong? If yes, is it a proof?
 A: Your argument does not work in general. See Mohan's comment for example. However, it does work if the following two conditions are satisfied:
i) The point $0\in\mathbb{R}$ is a regular value of $f_s=f+sm$ for every $s\in[0,1]$.
ii) The fiber $f_s^{-1}(0)$ (which is smooth, as follows from condition i)) is compact for every $s\in[0,1]$.
To see this, we shall use the following Fact: If $\pi:M\to N$ is a submersion between two smooth manifolds, and if the fiber $\pi^{-1}(n)$ is compact for every $n\in N$, then all the fibers are diffeomorphic to one another.
We now prove that under the above conditions, all the sets $f_s^{-1}(0)$ are diffeomorphic to one another. Define $$M=\{(x_1,\ldots,x_n,s)\in\mathbb{R}^n\times[0,1]|f_s(x_1,\ldots,x_n)=0\}.$$As $0$ is a regular value of $f_s$ for every $s$, it is surely a regular value of $$\phi:\mathbb{R}^n\times[0,1],\quad(x_1,\ldots,x_n,s)\mapsto f_s(x_1,\ldots,x_n).$$Consequently, $M$ is a smooth manifold (with boundary at $s=0,1$). Furthermore, at every $p\in M$ we have $$T_pM=\ker d\phi_p\not\subset T_p\mathbb{R}^n$$(keeping in mind the decomposition $T_pM=T_p\mathbb{R}^n\oplus T_p[0,1]$). This means that the projection$$\pi:M\to[0,1],\quad(x_1,\ldots,x_n,s)\mapsto s$$is a submersion. Its fibers are compact by condition ii) and hence diffeomorphic to one another by the quoted fact.
