Transversality of two mappings and diagonal I've never taken differential topology and am confused by the definition of transversality, and while trying to solve the following I got stuck.

Given smooth manifolds and maps $f:M\to N$ and $g:P\to N$, show that $f$ and $g$ are transversal to each other if and only if $f\times g: M\times P\to N\times N$ is transversal to the diagonal $\Delta \subset N\times N$.

 A: By definition $f \pitchfork g$ means that for every $x \in M$ and $y \in P$ such that $z\doteq f(x) = g(y)$, we have $${\rm d}f_x[T_xM] + {\rm d}g_y[T_yP] = T_zN.$$This is not required to be a direct sum. If $S \subseteq N$ is a submanifold, we write $f \pitchfork S$ instead to mean $f \pitchfork \iota_S$, where $\iota_S\colon S \hookrightarrow N$ is the inclusion map.
It is easy to see that ${\rm d}(f\times g)_{(x,y)}[M\times P] = {\rm d}f_x[T_xM] \times {\rm d}g_y[T_yP]$. Also, we have that the tangent space to the diagonal is the diagonal of the tangent space, i.e., $T_{(z,z)}\Delta = \{ (w,w) \mid w \in T_zN\}$.
$\implies:$ Assume that $f \pitchfork g$, and let's prove that $(f\times g) \pitchfork \Delta$. Assume that $(x,y) \in M\times P$ are such that $z \doteq f(x) = g(y)$ and take $(w_1,w_2) \in T_{(z,z)}(N\times N) = T_zN\times T_zN$. Our goal is to write $(w_1,w_2)$ as the sum of something in ${\rm d}f_x[T_xM]\times {\rm d}g_y[T_yP]$ with something in $T_{(z,z)}\Delta$. Well, $f \pitchfork g$ gives $u \in T_xM$ and $v\in T_yP$ such that $w_1-w_2 = {\rm d}f_x(u) - {\rm d}g_y(v)$. Now let $w = w_1 - {\rm d}f_x(u)$ and note that $w = w_2 - {\rm d}g_y(v)$ holds as well. Then we have that $$(w_1,w_2) = ({\rm d}f_x(u),{\rm d}g_y(v)) + (w,w),$$and this shows that $(f\times g)\pitchfork \Delta$, as wanted.
$\impliedby:$ Assume that $(f\times g)\pitchfork \Delta$ and let's show that $f \pitchfork g$. So, take $w \in T_zN$ and look at $(w,0) \in T_{(z,z)}(N\times N)$. The assumption $(f\times g)\pitchfork \Delta$ provides $u \in T_xM$, $v \in T_yP$ and $w' \in T_zN$ such that $$(w,0) = ({\rm d}f_x(u), {\rm d}g_y(v)) + (w',w').$$It readily follows that $w = {\rm d}f_x(u)+{\rm d}g_y(-v)$, which shows that $f \pitchfork g$.
