# Example of transversal map f where boundary of preimage is not preimage intersect boundary

I am utterly confused by this example in Guillemin and Pollack gave in "differential topology" page 60.

For a manifold with boundary, We would like conditions that would guarantee that if $f: X \rightarrow Y$ encounters a submanifold Z of Y, then $f^{-1}(Z)$ is a manifold with boundary. We also want $\delta f^{-1}(Z) = f^{-1}(Z) \cap\delta X$. Unfortunately, the transversality of f alone doesn't guarantee this. (For example, let $f: H^2 \rightarrow R$ be the map $(x_1, x_2) \rightarrow x_2$, and let Z be ${0}$. Then $f^{-1}(Z) = \delta H^2$.

How does the example they gave not satisfiy the condition $\delta f^{-1}(Z) = f^{-1}(Z) \cap\delta X$?

• What does $H^2$ denote? – Lee Mosher May 31 '18 at 3:18
• Sorry, $H^2$ denote the upper half space of $R^2$ – Ecotistician May 31 '18 at 6:15

Ok, So the reason why this does not satisfy the condition $\delta f^{-1}(Z) = f^{-1}(Z) \cap \delta X$ is because $\delta H^2$ by itself is a 1 manifold without boundary, so its boundary is actually the empty set, hence it is certain not equal to $\delta f^{-1}(Z) = f^{-1}(Z) \cap \delta X = \delta H^2$ in this case.