If $p$ is homogeneous of order $k$, then $p^{-1}(a)$ is diffeomorphic to $p^{-1}(b)$ Given homogeneous function $p$ with order $m$, how can I show that
$$p^{-1}(a) = \left(\frac{a}{b}\right)^{\frac{1}{m}}p^{-1}(b)?$$
The original question is:

Let $p$ be any homogeneous polynomial in $k$-variables. Homogeneity means 
  $$p(tx_1, \dots, tx_k) = t^mp(x_1, \dots, x_k).$$ 
  Prove that the set of points $x$, where $p(x) = a$, is a $k-1$ dimensional submanifold of $\mathbb{R}^k$, provided that $a \neq 0$. Show that the manifolds obtained with $a > 0$ are all diffeomorphic, as are those with $a <0$.

So I am wondering, is $m$ is odd, $a,b$ does not need to have the same sign to be homegeneous, since $\lambda = \left(\frac{b}{a}\right)^{1/m}$ still make sense?
Thanks.
 A: We are given that $p : \Bbb{R}^k \to \Bbb{R}$ is a homogeneous function of order $m$, and we have that $p^{-1}(a) = \{ x \in \Bbb{R}^k : p(x) = a \}$ by the definition of $p^{-1}$. Now, if $a, b > 0$, then let $t = (b/a)^{1/m}$. Suppose $x \in p^{-1}(a)$. Then $p(x) = a$, which implies that $$p(tx) = t^m p(x) = (b/a) \cdot a = b.$$ Hence, $tx \in p^{-1}(b)$. Conversely, suppose $x \in \Bbb{R}$ such that $tx \in p^{-1}(b)$. Then $p(tx) = b$. But, $p(tx) = t^m p(x) = (b/a) \cdot p(x)$. Hence, $$p(tx) = b \implies (b/a) \cdot p(x) = b \implies p(x) = a.$$
Hence, $x \in p^{-1}(a)$.
Thus, we have shown that $x \in p^{-1}(a) \iff tx \in p^{-1}(b)$, where $t = (b/a)^{1/m}$. This can also be expressed as
$$
p^{-1}(a) = (a/b)^{1/m} p^{-1}(b),\tag{$*$}
$$
where it is understood that
$$
(a/b)^{1/m} p^{-1}(b) := \{ (a/b)^{1/m} x : x \in p^{-1}(b) \}.
$$

Note that we have to assume that $m \neq 0$ in the question, otherwise $p^{-1}(a)$ is either empty or all of $\Bbb{R}^k$, neither of which are $(k-1)$-dimensional submanifolds of $\Bbb{R}^k$.
Regarding the additional question about whether or not $m$ is odd, the answer is no, there is no such restriction on $m$. You are asked to show that $p^{-1}{a}$ and $p^{-1}(b)$ are diffeomorphic when $a$ and $b$ are both positive and when $a$ and $b$ are both negative. When they have opposite sign they may or may not be diffeomorphic, but the above analysis does not answer this question (and that is also not part of the original problem in your post). I am not entirely sure what you mean when you say

$a,b$ does not need to have the same sign to be homegeneous, since $\lambda = \left(\frac{b}{a}\right)^{1/m}$ still make sense?

My guess is as above, that you are wondering about whether $p^{-1}(a)$ and $p^{-1}(b)$ can be diffeomorphic when $a$ and $b$ have opposite sign. If you meant something else then I would be happy to hear any clarifications in the comments.

Lastly, there seems to be some confusion regarding the notation $p^{-1}$. It is not a function, so there is no meaning to the question of whether $p^{-1}(a)$ is unique. (I suppose you mean to ask whether it is well-defined?) Note that ($*$) is an equality of sets, not of real numbers; this might possibly be the source of your misconception.
A: According to @Daniel Fischer's kind advice:
$$\vec{v} \in p^{-1}(a)$$
$$\lambda \vec{v} \in p^{-1}(\lambda^m a) \Rightarrow \vec{v} \in \frac{1}{\lambda} p^{-1}(\lambda^m a)$$
So assuming the preimage is unique, we have
$$p^{-1}(a) = \frac{1}{\lambda} p^{-1}(\lambda^m a)$$
Set $\lambda = \big(\frac{b}{a}\big)^{\frac{1}{m}}$, we get
$$p^{-1}(a) = \Big(\frac{b}{a}\Big)^{-\frac{1}{m}} p^{-1}\Big(\Big(\frac{b}{a}\Big)^{\frac{1}{m} \cdot m}a\Big) \Rightarrow p^{-1}(a) = \Big(\frac{a}{b}\Big)^\frac{1}{m} p^{-1}(a)$$
Though, I'm not certain about assuming uniqueness.
