Tangent Space to Fiber Product of Transverse Mappings

Question

I'm trying to show two things: that for two smooth transverse mappings $f:X \to Z$ and $g:Y \to Z$, the map $h:W \to Z\times Z$, $h'(x,y)(v_x,v_y) = (f'(x)(v_x), g'(y)(v_y))$ is surjective (I see that it happens to be the case that this is true for vector spaces, but I'm frustrated that I can't prove it in general), and the tangent space $T_{(x,y)}W$ where $W$ consists of $(x,y) \in X \times Y$ such that $f(x) = g(y)$ is contained in the kernel of the map $\phi:T_xX \times T_yY \to T_zZ$ defined by $\phi(v_x,v_y) = f'(x)(v_x) - g'(y)(v_y)$. Any help would be appreciated.

Answer 1

The idea in your last comment works indeed. And local is enough, as being an embedded submanifold is a local property. That is, it is enough to show that for any $(x,y)\in X\times Y$ with $f(x)=g(y)$, there is an open neighborhood $(x,y)\in U\subset X\times Y$ such that $W\cap U$ is an embedded submanifold of $U$.

Now, suppose $f(x_0)=g(y_0)=z_0$, and let $(V,\phi)$ be a coordinate chart of $Z$ around $z_0$. Write $U:=f^{-1}(V)\times g^{-1}(V)\subset X\times Y$, and define$$k:U\to\mathbb{R}^k,\quad (x,y)\mapsto \phi(f(x))-\phi(g(y)).$$ By construction, we have $W\cap U=k^{-1}(0).$ It follows from transversality that $0$ is a regular value of $k$, and the claim follows from the implicit function theorem.