I'm reading Voisin's proof of Ehresmann's lemma, given in the book Hodge Theory and Complex Algebraic Geometry I as theorem 9.3; the proof is essentially reproduced in this answer.
To summarize, there's a surjective proper submersion $F:M\rightarrow N$, and the theorem shows $F$ is a fiber bundle.
To do that, it looks at an arbitrary $q\in N$ and the compact embedded submanifold $S=F^{-1}(q)\subset M$. Using a tubular neighborhood $S\subset U$ with its associated retraction $r:U\rightarrow S$, a map $\Phi:U\rightarrow S\times N$ is defined by \begin{equation*} \Phi(p)=(r(p), F(p)) \end{equation*}
To finish the proof, first $U$ is shrunk down to an open neighborhood $S\subset V\subset U$ such that $\Phi:V\rightarrow S\times Y$ is an embedding. Then using properness an open neighborhood $q\in W\subset N$ is found with $F^{-1}(W)\subset V$.
I understand all of the above. It's the last step that I can't figure out: \begin{equation*} \Phi\left(F^{-1}(W)\right)=S\times W \end{equation*}
Why is this an equality? I've tried using "vertical slices": taking $y\in W$ and showing the retraction $r:F^{-1}(y)\rightarrow S$ is surjective. But I can only show it's a local diffeomorphism. I've also tried using "horizontal slices": trying to show for every $x\in S$ that $W\subset F(r^{-1}(x))$, but I've also failed to do that.
Both proofs start this step with "Clearly", so I'm wondering if I'm missing something obvious?