# Understanding submersions in differential topology

I'm having trouble understanding an example in Guillemin and Pollack. If $$f:R^k\rightarrow R$$ be defined by

$$f(x) = |x|^2 = x_1^2+...+x_k^2$$

The derivative $$df_x$$ at the point $$a = (a_1,..,a_k)$$ has matrix $$(2a_1,..,2a_k)$$. Thus

$$df_a:R^k\rightarrow R$$ is surjective unless $$f(a)=0$$, so every nonzero real number is a regular value of $$f$$. I can't understand how to check $$df_a$$ is surjective and why it is obvious that every nonzero real number is a regular value of $$f$$? A hint is appreciated. Thanks.

It is clear that if $$f(a)=a_1^2+\cdots+a_k^2 = 0$$, then $$a=0$$ which implies $$df_a = (0,\dots,0)$$ is the zero map (obviously not surjective).

If $$f(a)\neq 0$$, then at least there is one component of $$a$$ which is nonzero. Wlog, assume $$a_1 \neq 0$$. Therefore the components of $$df_a$$ not all zero. For any nonzero $$x \in \Bbb{R}$$, $$df_a([x/2a_1, \, 0, \cdots ,\, 0]) = 2a_1 \frac{x}{2a_1} + 2a_2 \cdot 0 + \cdots + 2a_k \cdot 0 = x.$$ So $$df_a$$ is onto.

Since any nonzero real number is the image of a particular vector under all linear map $$df_a$$ (where $$a$$ is point such that $$f(a) \neq 0$$), therefore they're all regular value.

Hint: What is the rank of the linear map $$2(a_1,\dots, a_k):\Bbb R^k\rightarrow \Bbb R$$? (And, what is the dimension of $$\Bbb R$$?)

• $dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements? – manifolded Jan 31 at 8:56
• The rank always equals the dimension of the column space, which equals that of the row space. So...? – Chris Custer Jan 31 at 9:01

For $$a \neq 0$$, the reason why $$df_a$$ is surjective because $$df_a$$ is a linear map from $$T_p \mathbb{R}^k$$ to $$T_{f(p)} \mathbb{R}$$ and since $$\operatorname{rank} df_a = 1 = \operatorname{dim} T_{f(p)}\mathbb{R}$$ it follows that $$df_a$$ is surjective.

To see that $$\operatorname{rank} df_a = 1$$, note that the matrix $$[ 2a_1 \dots 2a_k]$$ will always have some entry to be non-zero (since we assumed that $$a \neq 0$$ so some $$a_i$$ must be non-zero for some $$0 \leq i \leq k$$) and having one non-zero entry is all we need to conclude that the rank of this matrix is $$1$$.

Recall that the for a smooth map $$F : M \to N$$ between smooth manifolds, a point $$c \in N$$ is a regular value of $$F$$ if and only if either $$c$$ is not in the image of $$F$$ or at every point $$p \in F^{-1}(c)$$ we have $$dF_p : T_pM \to T_{F(p)}M$$ to be surjective.

So choose some non-zero real number $$y$$, if $$y \not\in f[\mathbb{R^k}]$$ then we're done otherwise suppose that $$y \in f[\mathbb{R^k}]$$, then $$f^{-1}(y)$$ is nonempty. Choose any $$a \in f^{-1}(y)$$, and then note that $$a$$ must clearly be non-zero in $$\mathbb{R}^k$$, thus by the above we have $$df_a$$ to be surjective and since $$a$$ was chosen arbitrarily here, this holds for all $$a \in f^{-1}(y)$$, and so it follows that every non-zero real number is a regular value for $$f$$.