Suppose that $f: X \rightarrow Y$ and $Z$ is a submanifold of $Y$, all boundaryless. Suppose that $f$ is transverse to $Z$, so that: $$df_x T_xX + T_{f(x)}Z = T_{f(x)}Y$$ for every $x \in f^{-1}(Z)$ and $codim(f^{-1}(Z)) = codim (Z)$. The manifolds are also "appropriate for intersection theory" in the sense that $dim(X) + dim(Z) = dim(Y)$.
Is it true that all such $f$ are local diffeomorphisms on $x \in f^{-1}(Z)$?
From what I understand, $df_x T_x X$ has dimension at most $dim(X)$, and the other two tangent spaces have exactly $dim(Z)$ and $dim(Y)$ respectively. Given the condition on dimensions, it seems that we must have $df_x T_x X$ be of dimension $dim(X)$, so that $df_x$ is a full-rank map. Thus, $df_x$ must be an isomorphism, and from the inverse function theorem, $f$ must be a local diffeomorphism at $x$.
I'm hoping that someone could confirm this, or point out any errors in the reasoning if untrue.
For general dimensions a standard transversality result (It is just analysis and can be found in Hirsch or Bröcker Jänich) gives you that $im(f)\cap Z$ is a smooth submanifold of dimension $dimX+dimZ - dim Y$, i.e. the pullback $M$ of the maps $f$ and the inclusion of $Z$ stays in category:
$$ \begin{array}{c} M& \hookrightarrow &Z\\ \downarrow &&\downarrow\\ X & \to &Y \end{array} $$
So we have that $X$ restricted to $f^{-1}Z=M \subset X$ is an embedding, i.e. a diffeomorphism onto its image $M\hookrightarrow Y$.
In the case that we consider complementary dimensions, the intersection will be 0-dimensional, i.e. a bunch of points. In this case it just means, that the points, will be full rank points of $f$ as you noticed correctly, since otherwise by linear algebra we wouldn't add up to the right dimension.
But it does not mean that it is an isomorphism, since we still could have $dimX < dim Y$ or equivalently $dimZ\neq 0$!