# Why the matrix of $dG_0$ is $I_l$.

I am reading the proof of Local Immersion Theorem in Guillemin & Pallock's Differential Topology on Page 15. But I got lost at the following statement:

Define a map $G: U \times \mathbb{R}^{l-k} \rightarrow \mathbb{R}^{l}$ by $$G(x,z) = g(x) + (0,z).$$

The matrix of $dG_0$ is $I_l$.

Could some one explain why this is true? Thanks.

Earlier on that page, it is said that since $dg_0: \Bbb R^k \longrightarrow \Bbb R^l$ is injective, we may assume that $$dg_0 = \begin{pmatrix} I_k \\ 0_{k-l} \end{pmatrix}$$ by performing a change of basis. Hence we have that the map $$\tilde{g}: U \times \Bbb R^{l-k} \longrightarrow \Bbb R^l,$$ $$\tilde{g}(x, z) = (g(x), 0, \dots, 0)$$ has differential $$d\tilde{g}_0 = \begin{pmatrix} I_k & 0_{l-k} \\ 0_{l-k} & 0_{l-k} \end{pmatrix}.$$ The map $$h: U \times \Bbb R^{l-k} \longrightarrow \Bbb R^l,$$ $$h(x,z) = (0, \dots, 0, z)$$ clearly has differential $$dh_0 = \begin{pmatrix} 0_{k} & 0_{l-k} \\ 0_{l-k} & I_{l-k} \end{pmatrix}.$$ Therefore the given map $$G: U \times \Bbb R^{l-k} \longrightarrow \Bbb R^l,$$ $$G(x,z) = \tilde{g}(x ,z) + h(x,z)$$ has differential $$dG_0 = d\tilde{g}_0 + dh_0 = \begin{pmatrix} I_{k} & 0_{l-k} \\ 0_{l-k} & I_{l-k} \end{pmatrix} = I_l.$$
• Hi Henry, I tried a long time but still not able to figure out why $dg_o$ is injective. Could you give me a hint..? Thanks! – WishingFish Jun 30 '13 at 21:23
• @WishingFish How is the differential of the maps h and $\bar g$ calculated? – Idonotknow Oct 10 '18 at 21:07