Transversality

Remarks

The condition is empty where $f(X)$ misses $Z$, so a map whose image avoids $Z$ is transverse to $Z$. If $Z$ is a point $y$, then $f \pitchfork \set{y}$ means that $y$ is a regular value of $f$. If $f \pitchfork Z$, then $f^{-1}(Z)$ is an embedded submanifold of $X$ with $\dim X - \dim f^{-1}(Z) = \dim Y - \dim Z$. Example: in $\R^2$ the $x$-axis and the line $y = x$ are transverse at the origin, since their directions $(1, 0)$ and $(1, 1)$ span $\R^2$. Non-example: the $x$-axis and the parabola $y = x^2$ meet at the origin, where both tangent lines equal the $x$-axis, which does not span $\R^2$; so they are not transverse.