Additivity of Codimension in Transversality

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In Guillemin and Pollack's Introduction to Differential Topology, the following theorem is given:

Theorem. The intersection of two transversal submanifolds $X,Z$ of $Y$ is again a submanifold. Moreover,, $\operatorname{codim}(X \cap Z) = \operatorname{codim}X + \operatorname{codim}Z$.

The statement of the theorem is followed by a short discussion on the intuitive nature of the additivity of codimension:

The additivity of codimension...is absolutely natural. Around a point $x$ belonging to $X \cap Z$, the submanifold $X$ is cut out by $k = \operatorname{codim}X$ independent functions, and $Z$ is cut out by $\ell = \operatorname{codim}Z$ independent functions. Then $X \cap Z$ is locally just the vanishing set of the combined collection of $k + \ell$ functions; that these $k + \ell$ functions are together independent is exactly the transversality condition [$T_x(X) + T_x(Z) = T_x(Y)$].

I'm having trouble proving the bolded part for myself. I know that if $g = (g_1,\ldots,g_k) : Y \to \mathbb{R}^k$ are the independent functions that cut out $X$ and $h = (h_1,\ldots,h_\ell) : Y \to \mathbb{R}^\ell$ are the independent functions that cut out $Z$, then $\operatorname{ker}(dg_x) = T_xX$ and $\operatorname{ker}(dh_x) = T_xZ$, but I don't know where to go from here.

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There are various ways to look at this, but I think the simplest is to put together two things from linear algebra: (1) the nullity-rank theorem, and (2) the formula for the dimension for the sum of two subspaces.

Let $S\colon\Bbb R^n\to \Bbb R^k$ and $T\colon\Bbb R^n\to\Bbb R^\ell$ be surjective linear maps. Then $(S,T)\colon\Bbb R^n\to\Bbb R^{k+\ell}$ is surjective if and only if $\dim\ker(S,T) = n-(k+\ell)$. Of course, $\ker(S,T) = \ker S\cap \ker T$. (In our application, $\ker S$ is the tangent space of one submanifold and $\ker T$ is that of the other.) Now when do we have $\ker S + \ker T = \Bbb R^n$? Well, \begin{align*} \dim(\ker S+\ker T) &= \dim\ker S + \dim\ker T - \dim(\ker S\cap \ker T) \\&= (n-k) + (n-\ell) - \dim(\ker S\cap \ker T) \\&= n \end{align*} if and only if $$\dim(\ker S\cap\ker T) = n-(k+\ell),$$ i.e., if and only if $(S,T)$ is surjective.