The preimage of $\triangle$ is a compact zero-dimensional manifold. $f: X \to Y$, $g: Z \to Y$ and $Z$ are appropriate for intersection theory ($X,Y,Z$ are boundaryless oriented manifolds, $X,Z$ is compact, $Z$ is closed submanifold of $Y$, and $\dim X + \dim Z = \dim Y$), $f$ is transversal to $Z$.

(1) If $\triangle$ denotes the diagonal of $Y \times Y$, and $f \times g : X \times Z \to Y \times  Y$ is the product map, then $f(x) = g(z)$ precisely at pairs $(x,y)$ in $(f \times g)^{-1}(\triangle)$. Prove $\dim (X \times Z) = \operatorname{codim} \triangle$.
(2) If $f \times g \pitchfork \triangle$, the preimage of $\triangle$ is a zero-dimensional manifold.
(3) The preimage of $\triangle$ is compact.

 A: Jellyfish, you need to pay attention to details yourself! Aren't $X$ and $Z$ complementary dimension in $Y$? Do the arithmetic. No, it doesn't assume transversality of the maps.
A: Following Ted Shifrin's guidance, here's my own work out for this problem:
(1) Since $\dim \triangle = Y$, we easily get $\dim(X \times Z) = \dim X + \dim Z = \dim Y = \dim Y \times Y - \dim \triangle = \operatorname{codim} \triangle$.
(2) According to the theorem:

Theorem If the smooth map $f: X \to Y$ is transversal to a submanifold $Z \subset Y$, then the preimage $f^{-1}(Z)$ is a submanifold of $X$. Morever, the condimension of $f^{-1}(Z)$ in $X$ equals the codimension of $Z$ in $Y$.

Hence, for this question, we have we have  $f \times g : X \times Z \to Y \times  Y$, so the codimension of $(f \times g)^{-1}(\triangle)$ equals the codimension of $\triangle$ in $Y \times Y$, which is zero dimensional.
(3) The preimage of $\triangle$ is the subset of a compact set $X$, hence the preimage of $\triangle$ is compact. Hence, since the domain $X$ is compact, the preimage of $\triangle$ is compact.
