How to show $\dim T_x(X) = \dim df_x T_x(X).$

Question

$f: X \to Y$ and $Z$ are appropriate for intersection theory ($X,Y,Z$ are boundaryless oriented manifolds, $X$ is compact, $Z$ is closed submanifold of $Y$, and $\dim X + \dim Z = \dim Y$), $f$ is transversal to $Z$. According to the text:

If $f(x) = z \in Z$, then transversality plus dimensional complementarity give a direct sum $$df_xT_x(X) \oplus T_z(Z) = T_z(Y). \tag{1}$$ - Guillemin and Pollack, Differential Topology Page 108

So I am missing a piece here. From transversality, I know $$df_xT_x(x) + T_z(Z) = T_z(Y). \tag{2}$$

From dimensional complementarity, I know $$\dim X + \dim Z = \dim Y,$$ and by linearity of tangent space, I know $$\dim T_x(X) + \dim T_y(Z) = \dim T_y(Y). \tag{3}$$

But in order to get $(1)$ from $(2)(3)$, I need $\dim T_x(X) = \dim df_x T_x(X).$

Can anyone fill this up for me?

Also, consequently,

The orientation of $X$ provides an orientation of $df_xT_x(X).$ Then the orientation number at $x$ is $+1$ if the orientation on $df_xT_x(X)$ and $T_z(Z)$ "add up" to the prescribed orientation on $Y$.

So I am totally confused. What does it mean by "add up" to the prescribed orientation?

Definition: Orientation of $V$, a finite-dimensional real vector space: Let $\beta, \beta^\prime$ be ordered basis of $V$, then there is a unique linear isomorphism $A: V \to V$ such that $\beta = A \beta^\prime$. The sign given an ordered basis $\beta$ is called its orientation.

Definition: Orientation of $X$, a manifold with boundary: A smooth choice of orientations for all the tangent space $T_x(X).$

Answer 1

Regarding the first part, for every linear map $\Lambda \colon V \to W$, you have

$$\dim V = \dim \ker \Lambda + \dim \operatorname{im} \Lambda,$$

as can be seen by extending a basis of $\ker \Lambda$. Consequently, $\dim \operatorname{im} \Lambda \leqslant \dim V$.

So you have

$$\begin{align} T_z(Y) = T_z(Z) + df_x T_x() &\Rightarrow \dim df_x T_x(X) \geqslant \dim Y - \dim Z = \dim X\\ \dim df_x T_x() = \dim X - \dim \ker df_x &\Rightarrow \dim df_x Tx(X) \leqslant \dim Y - \dim Z = \dim X \end{align}$$

together, that implies $\dim df_x T_x(X) = \dim T_x(X) = \dim X$, and therefore $df_x T_x(X) \cap T_z(Z) = \{0\}$, hence the sum is direct.

Regarding the second, $df_x$ maps a basis belonging to the orientation of $X$ at $x$ to a basis $\mathcal{B}_X = (v_1,\, \dotsc,\, v_k)$ of $df_x Tx(X) \subset T_z(Y)$. You then pick a basis $\mathcal{B}_Z = (w_1,\, \dotsc,\, w_m)$ of $T_z(Z) \subset T_z(Y)$ belonging to the orientation of $Z$ at $z$. You then combine the two bases to a basis $(v_1,\, \dotsc,\, v_k,\, w_1,\, \dotsc,\, w_m)$ of $T_z(Y)$ - it is a basis because you have a direct sum decomposition of $T_z(Y)$. That's how the two orientations "add up". If the combined basis belongs to the orientation, the orientation number is $+1$ else $-1$.

Answer 2

From transversality (2), you know $\dim f_xT_x(X)\geq \dim Y-\dim Z$. From linear algebra, you know that the image of a linear map has at most the dimension of the domain, so $\dim f_xT_x(X)\leq\dim X$. Combine those two inequalities with the complementarity equation (3) to see that both inequalities are in fact equalities, and you have what you needed.