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$f: X \to Y$, $g: Z \to Y$ 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.

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1LiterTears
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2 Answers2

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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.

Ted Shifrin
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  • I'm sorry Ted, I overlooked the fact that $X$ and $Z$ are complementary dimension.... I will be more careful and read back and forth in the future... – 1LiterTears Jul 22 '13 at 20:10
  • But I got stuck immediately again at the later half of the sentence..... I have tried a lot of work out, but it seems not trivial(and I can't work through....) – 1LiterTears Jul 22 '13 at 22:43
  • Thanks again Ted, I hope I solved it eventually, as updated answer below.. – 1LiterTears Jul 23 '13 at 17:00
  • Not quite yet. (1) needs to say $\text{codim},\Delta$. In (2), "which is zero dimensional" is wrong. And in (3), preimage of compact needn't be compact! How do you fix this? – Ted Shifrin Jul 23 '13 at 18:12
  • Dear Ted, thanks so much for your kindness. (1) is a mistake and I corrected it, hopefully. (2) I understand your point but don't know how to fix it. (3) I meant to say, the preimage of $\triangle$ is a subset of $X$. Since $X$ is compact, then the preimage of $\triangle$ is bounded. Yeah, I haven't figure out the trick to show it is closed......... – 1LiterTears Jul 23 '13 at 20:15
  • What is the codimension of $\Delta$ in $Y\times Y$? And surely you've taken basic analysis before you're trying to do Guillemin and Pollack! Review your basics about open and closed sets. – Ted Shifrin Jul 23 '13 at 20:34
  • I thought the codimension of $\triangle$ in $Y \times Y$ is $Y$? – 1LiterTears Jul 23 '13 at 23:23
  • Codimension of $N$ in $M$ is, by definition, $\dim M -\dim N$. You have got to learn definitions and pay attention to detail. – Ted Shifrin Jul 23 '13 at 23:27
  • Yes, I am aware of that. Is $\dim Y \times Y = 2 \dim Y$, and $\dim \triangle = \dim Y$, hence the codimension of $\triangle$ in $Y \times Y$ is $2\dim Y - \dim Y = \dim Y$? – 1LiterTears Jul 23 '13 at 23:29
  • Ok. Please proofread everything carefully here in the future before expecting us to make sense of it! – Ted Shifrin Jul 23 '13 at 23:33
  • OK, I'll be more careful, Ted.. – 1LiterTears Jul 23 '13 at 23:35
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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.

1LiterTears
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  • So, if $f\times g\pitchfork\Delta$, what dimension does $(f\times g)^{-1}(\Delta)$ have? You need to use theorems, not try to go back to definitions every time. – Ted Shifrin Jul 22 '13 at 23:08
  • Thanks good Professor @TedShifrin~ – 1LiterTears Jul 22 '13 at 23:10
  • Is it preimage theorem? I managed to show each $(y,y) \in Y \times Y$ is a regular value of $f \times g$: $g \times f \pitchfork \triangle \Leftrightarrow d (g \times f)$ is surjective $\Leftrightarrow$ satisfies preimage theorem $\Leftrightarrow \dim (f \times g)^{-1} = \dim Y - \dim X - \dim Z =0.$ Since each $(y,y)$ has dimension 0, so does altogher?.. – 1LiterTears Jul 22 '13 at 23:59
  • Reread pp. 27-28 of G&P. The whole point of transversality is the theorem at the bottom of p. 28. – Ted Shifrin Jul 23 '13 at 00:02
  • That's really, really, very, very helpful Ted. Thank you so much. However, may I ask if my approach in the comment above is correct..... seems very shaky to myself... @TedShifrin – 1LiterTears Jul 23 '13 at 00:12
  • First of all, you're at a point $(y,y)\in\Delta$, so no $y_1,y_2$. It is true that the rank of $f\times g$ at $(x,z)$ must be at least $=\dim Y$ in order to have transversality to the diagonal, and so equality must in fact hold and each of $f$ and $g$ must be an immersion. But you're making basic errors here: Looking at dimensions of vector spaces is NOT good enough; the particular vector spaces must add up to $T_{(y,y)}(Y\times Y)$, and dimensions alone will only tell you that this fails, never that it succeeds. – Ted Shifrin Jul 23 '13 at 00:54