I am just learning about the concept of transversality so please be patient with me. So far I understand that a map $f$ is transversal to a submanifold $Z$ if $\text{Im}(d_xf) + T_yZ = T_yY$. Now I consider this example: $f:\mathbb{R} \to \mathbb{R}^2$, $x \mapsto (0,x)$ and $Z$ is the x-axis.

Now I try to construct the tangent spaces (also this is kind of new to me). The tangent space of $f$ should be the map $(0,v)$ for $v \in \mathbb{R}$. But now I am stuck. The tangent space $T_yY$ should be $\mathbb{R}^2$ again, right? But as $Z$ is the x-axis isn't the tangent space for example at point $0$ just the y-axis? So I don't see how I get in the end $\mathbb{R}^2$. But maybe I made some mistakes in between?


Since $Y= \mathbb R^2$, you are correct that $T_y Y = \mathbb R^2$ for any $y\in \mathbb R^2$. As $Z$ is the $x$-axis, the tangent space contains all those vectors that are tangent to $Z$, thus for all $y=(x, 0) \in Z$, $T_yZ$ is also the $x$-axis.

You are sort of correct concerning the "tangent space of $f$. Indeed you need to find $\text{Im} (d_y f)$ instead of the tangent space of $\mathbb R$. For each $t\in \mathbb R$, the tangent map $d_t f : \mathbb R^2 \to T_{(f(t)} Y \cong \mathbb R^2$ is given by $v\mapsto (0,v)$. As a result, since $\text{Im} (f) \cap Z = \{y= (0,0)\}$ and at this point we have

$$ \text{Im} d_y f + T_y Z = \mathbb R^2,$$

thus $f$ is transversal to $Z$.

As an example, consider the mapping $g (t) = (0,t^3)$. Although this map has the same image as $f$ (that is, $f(\mathbb R) = g(\mathbb R)$), $g$ is not transvarsal to $Z$ since at $y = (0,0)$, the tangent map $d_0 g$ is zero.

  • $\begingroup$ Thanks! This makes a whole lot of sense to me. Only one thing I don't understand: Why do you stress that $T(f) \cap Z = \{(0,0)\}$? Because it is the only point where $f \cap Z \neq \emptyset$ and only there you have to check the transversality condition? $\endgroup$ – JDoe Aug 8 '16 at 18:46
  • $\begingroup$ It should be $\text{Im}$ instead of $\Im$, @JDoe Corrected. $\endgroup$ – user99914 Aug 8 '16 at 18:47
  • $\begingroup$ You are right, to check transvarsality you need only to check at the points in $\text{Im} f \cap Z$ (or the equation does not make sense) @JDoe $\endgroup$ – user99914 Aug 8 '16 at 18:51
  • $\begingroup$ Yeah I just realized this. To your example: This is the same reason why the map $g(x) = (x,x^2)$ is not transversal to map $Z$ correct? $\endgroup$ – JDoe Aug 8 '16 at 18:53
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    $\begingroup$ your $g$ is not transvarsal to $Z$, not because the tangent map at $0$ is zero, but because $d_0 g(\mathbb R) = T_y Z = \text{span} \{ (1,0)\}$ so your equality does not hold. @JDoe $\endgroup$ – user99914 Aug 8 '16 at 18:55

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