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I'm currently reading through the famous paper "Diffeomorphism of 4-manifolds" of C. T. C. Wall. It's farly understable, apart from a single point, where things seem to be a bit tricky:

"Assume $N$ (a compact, oriented 4-manifold without boundary) to be simply-connected. Then any two circles are homotopic, so —by the argument of Lemma 4 — isotopic, and so any one spans an imbedded 2-disc..."

Here is where I'm stuck. I get why every circle spans an immersed 2-disk having double points. But since $\dim(N) = 4 = 2\dim(\mathbb{D}^2)$, it seems to me that we cannot suppose such disks to be embedded. Right? If I remember correctly, the problem of embedding disks inside 4-manifolds is one of the reasons why $h$-cobordim smoothly fails in dimension 4.

Am I missing something?

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  • $\begingroup$ But how can we suppose the trace to be a proper submanifold? The fact that we have an isotopy $S^1 \times I -> N$ does not guarantee that it is itself an embedding $\endgroup$ Commented Aug 21 at 15:39
  • $\begingroup$ The question is what does mean by an isotopy: if it is an ambient isotopy then it is enough. $\endgroup$ Commented Aug 21 at 15:52
  • $\begingroup$ We can always suppose an isotopy to be ambient, correct? $\endgroup$ Commented Aug 21 at 16:07
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    $\begingroup$ It's easier to do the isotopy of the embedded circle first -- this is basically the 1-parameter version of the weak whitney embedding theorem. Once you've isotoped your embedded circle to be unknotted in a 4-ball, you have your spanning disc. This is a standard argument back in the 70's so that's likely why Wall didn't go into details -- there were dozens of papers at the time rewriting this argument in various different contexts. $\endgroup$ Commented Aug 21 at 16:12
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    $\begingroup$ Oh, of course. Then you get your embedded disk. $\endgroup$ Commented Aug 21 at 16:14

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Start with your smooth embedding

$f : S^1 \to N$.

Then you have a homotopy of $f$ to any other smooth embedding, let's write the homotopy as

$H : I \times S^1 \to N$

now you look at the track of the homotopy

$F : I \times S^1 \to I \times N$

this maps $F(t,p) = (t,H(t,p))$.

The domain is 2-dimensional and the co-domain is 5-dimensional so you can approximate this by a smooth embedding (weak whitney embedding).

A uniform approximation to a "tracked" smooth map is also a tracked map (apply the inverse/implicit function theorem correctly or quote the appropriate propostion from the Guillemin and Pollack text). So this approximation is the track of an isotopy.

That's how these arguments work. You can do Whitney approximations relative to embedded submanifolds, i.e. you don't have to perturb the parts that were embeddings to begin with.

Once you have your spanning disc for the "plain" embedded circle, you apply the Isotopy Extension Theorem to get the spanning disc for your original embedded circle.

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  • $\begingroup$ Thanks for the suggestions, but I still don't completely get it. Even if we have an embedding $f \colon I \times S^1 \to I \times N$, why does this induce an embedding $I \times S^1 \to N$? If we take $f$ to be the identity, it doesn't seem to work, at least to me. $\endgroup$ Commented Aug 21 at 16:39
  • $\begingroup$ Ahhh, I'm getting it. So what are you saying is the following: the "little unknot" has a stupid spanning disk. We know that it is homotopic to our circle $S^1 \hookrightarrow N$. But by Whintey, it is actually isotopic. Now, we can extend the isotopy to a global one. And we can "drag" our "little spanning disk" of the unknot to a spanning of $S^1$. Correct? $\endgroup$ Commented Aug 21 at 16:55
  • $\begingroup$ It doesn't induce an embedding of $I \times S^1 \to N$. The map $F$ you perturb to a tracked smooth embedding of $I \times S^1 \to I \times N$, i.e. that's the track of an isotopy. $\endgroup$ Commented Aug 21 at 16:55
  • $\begingroup$ Okay, sounds like the argument is making more sense now. $\endgroup$ Commented Aug 21 at 16:58
  • $\begingroup$ This argument gets used extensively to show things like the space of embeddings $Emb(M, \mathbb R^k)$ are highly-connected when $k$ is significantly larger than $dim(M)$. So the argument that $BDiff(M)$ is homotopy equivalent to $Emb(M, \mathbb R^\infty)/Diff(M)$ is one of the consequences. $\endgroup$ Commented Aug 21 at 17:27

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