# Transversality of maps

Let $X\xrightarrow{f} Y \xrightarrow{g} Z$ be a sequence of smooth maps of manifolds, and assume that $g$ is transverse to a submanifold $W$ of $Z$. Show $f\pitchfork g^{-1}(W) \leftrightarrow g\circ f \pitchfork W$.

From the definition of transverse, $$\forall x \in g^{-1}(W), \text{im}d_xg + T_ZW = T_ZZ$$

Then I was trying to start the first side of the proof, $f\pitchfork g^{-1}(W) \rightarrow g\circ f \pitchfork W$, but I'm confused about $f \pitchfork g^{-1}(W)$. Does it mean $$\forall x \in W, \text{im}d_xf + T_ZW = T_ZZ$$

I'm not sure where to go from there either.

No, first of all I'd suggest that you use letters to guide you to the locations of the points. Use $x$ for elements of $X$, $y$ for elements of $Y$, and $z$ for elements of $Z$. Next, your subscripts of $Z$ are incorrect. Maybe you want $x\in f^{-1}(g^{-1}(W))$, $f(x)=y$, and $g(y)=z$.
$f\pitchfork g^{-1}W$ means that for all $x\in f^{-1}(g^{-1}(W))$ we have $\text{im}(df_x) + T_{f(x)}(g^{-1}(W)) = T_{f(x)}Y$. But since $g\pitchfork W$, we have $T_y(g^{-1}(W)) = (dg_y)^{-1} T_{g(y)}W$. Note also that $(g\circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$. Can you make some progress now?
• How dod you know "But since $g\pitchfork W$, we have $T_y(g^{-1}(W)) = (dg_y)^{-1} T_{g(y)}W$." The definition of $g\pitchfork W$ tells us that $$\operatorname{Img}(dg_y) + T_z(W) = T_z(Z)$$ Commented Jan 24 at 18:03
• @Math_Day Two comments: (1) The standard argument is to reduce to the case of a regular value: Writing $W=\phi^{-1}(0)$ for $0$ a regular value, $g\pitchfork W \iff 0 \text{ is a regular value of } \phi\circ g$. (2) It is also useful to note that $g\pitchfork W$ if and only if $dg_y$ induces a surjective map $\overline{dg_y}\colon T_zZ/T_zW$. Commented Jan 25 at 4:02