# Showing transversality more rigorously.

In Guillemen and Pollack 2a asks if The $xy$-plane and the $z$-axis are transversal in $R^3$.

Intuitively I think the answer is yes as they "intersect" in a rather stable fashion. However I am having trouble verifying this more rigorously. To do so I am considering the definition of transversality:

$$\Im(df_x) + T_y(Z) = T_y(Y)$$

I considered taking $f : R^1 \to R^3$ to be the $z$ axis defined by:

$$f(t) = (0, 0, t)$$

So:

$$df_x = \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}$$

And so the image of $df_x$ is simply the span of this vector. But I am unsure what $T_y(Z) = T_y(R^2)$ and $T_y(Y) = T_y(R^3)$.

• I think you're on the right track. But the definition of "transversality" is not just the formula you cite: presumably there are some words before it. Like, "If $Y$ and $Z$ are ...., we say they are transverse if..." In other words, maybe you should write out the entire definition, including what $f$, $Y$, and $Z$ are. Commented Oct 1, 2016 at 21:05

Guilleman and Pollack discuss transversality for two submanifolds a page or so after giving the definition you cited. Namely, if $X,Z\subset Y$ are submanifolds of $Y$, then $X$ and $Z$ are transversal if and only if for all $y\in X\cap Z$, $$T_y(X)+T_y(Z)=T_y(Y).$$ The way this is seen is by taking the inclusion $\iota: X\to Y$, so $d\iota_y: T_y(X)\to T_y(Y)$ will just be the inclusion of tangent spaces.
The tangent space of the $z$-axis is just the $z$-axis. The tangent space of the $xy$-plane is just the $xy$-plane. Their sum will obviously span all of $\mathbb{R}^3$, so you're good.