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More generally, let $f: X \to Y$ be a map transversal to a submanifold $Z$ in $Y$. Then $W = f^{-1}(Z)$ is a submanifold of $X$. Prove that $T_x(W)$ is the preimage of $T_{f(x)}(Z)$ under the linear map $df_x:T_x(X) \to T_{f(x)}(Y)$.

("The tangent space to the preimage of $Z$ is the preimage of the tangent space of $Z$.") (Why does this imply The tangent space to the intersection is the intersection of the tangent spaces.?)

I am very lost at this question. My idea is that to show

$$T_x(W) = \{v \in T_x(X)\;|\;df_x(v) \in T_{f(z)}(Z)\}.$$

Not sure if this is the right track, and I have no clue how to proceed.

Any ideas? Thanks.

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1LiterTears
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2 Answers2

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@1LiterTears: You can try it yourself, you just need some definitions:

  1. $Z$ a submanifold of $Y$ of codimension $k$ if $Z$ is locally the preimage of zero by a submersion $\varphi : Y\to\mathbb{R}^k$, that is any $y\in Z$ admits a neighborhood $V$ (in $Y$) such that $D_y\varphi : T_yV=T_yY\to\mathbb{R}^k$ is surjective and $V\cap Z=\varphi^{-1}(0)$.

  2. For any $x\in f^{-1}(V\cap Z)=(\varphi\circ f)^{-1}(0)$, $$f\pitchfork Z\Longrightarrow x\ \text{is a regular point of}\ \varphi\circ f.$$

  3. The tangent space to a submanifold $Z$ in a point $y$ is $$T_yZ=\big\{\gamma'(0),\ \gamma:\,(-\varepsilon,\varepsilon)\to Z\ \text{smooth},\ \gamma(0)=y\big\}$$

With this you can prove that $T_yZ=\ker D_y\varphi$, $T_xW=\ker D_x(\varphi\circ f)=\ker D_{f(x)}\varphi\circ D_x f$. Then, $T_xW=(T_xf)^{-1}(T_{f(x)}Z)$.

Alex Ortiz
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amine
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Your idea doesn't make sense as written, since $df_x$ is defined on $T_x(X)$, not $X$ itself, and so $df_x(x)$ makes no sense (and neither does $df_w(w)$).

To answer this question, you will need to have a much stronger understanding of basic ideas like derivatives (i.e. the maps $df_x$) and tangent spaces than you currently seem to. Perhaps you should try some more basic questions first.

Also, this implies the result about tangent spaces of intersections pretty immediately: to say that submanifolds $X$ and $Y$ of $Z$ are transverse is the same as saying the the inclusion $\iota:X \to Z$ is transverse to $Y$. The preimage of $Y$ under $\iota$ is precisely $X \cap Y$. Now just apply the general statement in this particular case.

Alex Ortiz
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Matt E
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  • Oh this is just a question from the text. I thought they are easiest problems I could find.. – 1LiterTears Jul 09 '13 at 06:28
  • Can you suggest me some good source of practice for my level on this topic? – 1LiterTears Jul 09 '13 at 06:38
  • Yeah I see it is very much wrong. I shall be more careful in the future. I hope I made some correction now. – 1LiterTears Jul 09 '13 at 07:43
  • @Jellyfish: Dear Jellyfish, With the edited version of the question, the equation you have written down is just $T_x(W) =$ the preimage of $T_{f(x)}(Z)$. So while this is what you have to prove, it is just a restatement of the problem with more symbols. In other words, while it is a correct reformulation of the problem, it is nothing more than a reformulation. As for what problems to practice, I know that there are a lot of differential topology textbooks, but it's a long time since I looked at any, so I can't give any specific recommendations. I think that you should practice some ... – Matt E Jul 09 '13 at 19:14
  • ... basic problems, to get a better feel for the meaning of tangents spaces, $df_x$, etc., and in particular, to practice with some simple examples. At the moment, I get the impression that you are thinking entirely, or almost entirely, symbolically, and are not visualizing the situations you are asking about. (In particlar, transversality has a very distinct meaning when you visualize two submanifolds intersecting, and I'm not sure that you have grasped that meaning in a visual way yet.) Why don't you ask a question here on advice for beginning differential geometry texts, with good ... – Matt E Jul 09 '13 at 19:18
  • ... sets of problems in them? Regards, – Matt E Jul 09 '13 at 19:18
  • I have one question. Do we need transversality, apart from the fact that $W$ is a submanifold of $X$? – Dimitris Oct 22 '13 at 05:37
  • @DimitrisDallas: Dear Dimitris, Yes, we do. Consider the case when $X = Y = \mathbb R$, $Z = {0}$, and $f$ is the map $x \mapsto x^2$. Regards, – Matt E Oct 22 '13 at 06:03