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Take $S^n$ and consider the union $Z$ of $k_1$ circles, $k_2$ 2-dimensional spheres, ..., $k_{n-2}$ $(n−2)$-dimensional spheres, embedded in $S^n$ in an unknotted way, with no mutual intersection and no mutual linking.

I want to understand the complement $S^n \setminus Z$.

To that end, I calculated its homology via Mayer-Vietoris and I think that complement is homotopy equivalent to $(S^{n-1})^{k_1+k_2+\dots+k_{n-2}-1} \vee (S^{n-2})^{k_1} \vee \dots \vee (S^{1})^{k_{n-2}}$.

Is it true? Where can I find a reference? The problem is that we cannot deduce this isomorphism from knowing the homology... Is there any way to see this explicitly (if it's true)?

Originally asked as Math Stack Exchange question 3991178.

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    $\begingroup$ Alexander duality (en.wikipedia.org/wiki/Alexander_duality) is probably the easiest way of computing the homology of the complement. $\endgroup$ – Thomas Rot Jan 19 at 14:04
  • $\begingroup$ @ThomasRot Right. But I'm interested in a homotopy type of a complement rather than in homology itself. Is my suggestion about a homotopy type correct? Is it a wedge of spheres? $\endgroup$ – iou Jan 19 at 14:06
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    $\begingroup$ Alexander duality for manifolds can be refined to a statement about the Thom space of the normal bundle being stably equivalent to the complement of $X$ and a disjoint basepoint in the sphere. In fact, the stabilization need only happen once, and in your setting the stabilization has already happened since we can imagine these all embedded in a codimension 2 equator. The addition of the embeddings being uninteresting implies the normal bundles are trivial, so up to removing another point, yes the homotopy type is a wedge of spheres, probably not difficult to answer the rest from there. $\endgroup$ – Connor Malin Jan 19 at 14:48
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Your answer is correct, if "unknotted" is equivalent to: being homeomorphic to a union of standard sphere complements within disjoint convex balls. Philosophically, you could imagine giving a proof by saying that such a space is a connected sum of sphere complements, and understand enough about them to prove it inductively. Here is an argument that is a little more homotopy-theoretic, which calculates the answer inside a closed ball of large radius and then gets the answer for a sphere by gluing on a disc.

You need a basic calculation to get started. For $k \leq n-2$, the complement $C_k$ of the standard k-sphere (embedded in the first $(k+1)$ coordinates) in the standard n-ball of radius 2 has an explicit determination of its homotopy type. It has a deformation retraction down to the union of the (n-1)-sphere of radius 2, together with the $(n-k-1)$-dimensional ball of radius 2 in the last $(n-k-1)$ coordinates. (This is a "draw a straight line from the nearest point on the $k$-sphere" retraction, I believe.) This space formed by attaching an $(n-k-1)$-dimensional cell along a map $S^{n-k-2} \to S^{n-1}$; this is necessarily nullhomotopic, so the space is homotopy equivalent to $S^{n-1} \vee S^{n-k-2}$, as the homology suggests. Under this, the inclusion from the outside boundary is homotopic to the inclusion of the first wedge factor.

The complement $Y$ of $j$ nonintersecting convex balls in a closed ball of radius $R$ is homotopy equivalent to the complement of $j$ points in $R^n$, and this makes it homotopy equivalent to $\bigvee_{i=1}^{n} S^{n-1}$. We can use a degree calculation in homology to be more explicit: the inclusion of each boundary component of one ball is homotopic to the inclusion of a wedge factor, while the inclusion of the outside boundary is homotopic to the "pinch" map $S^{n-1} \to \bigvee S^{n-1}$.

The complement $Z$ of $k_1$ 1-spheres, $k_2$ 2-spheres, etc, in a closed ball of radius $R$ is the pushout: $$ Y \leftarrow \coprod S^{n-1} \rightarrow \coprod_i \left(\coprod^{k_i} C_{k_i}\right). $$ Up to homotopy equivalence, the previous identifications say that we have the pushout of $$ \bigvee^{\sum k_i} S^{n-1} \leftarrow \coprod^{\sum k_i} S^{n-1} \rightarrow \coprod \left(\coprod^{k_i} S^{n-1} \vee S^{n-1-k_i}\right), $$ with the map on the left collapsing basepoints and the map on the right being the coproduct of wedge inclusions. The result is that we identify basepoints in all the coproduct factors, and so the resulting space is homotopy equivalent to $$ \bigvee \left(\bigvee^{k_i} S^{n-1} \vee S^{n-1-k_i}\right) \cong \bigvee^{\sum k_i} S^{n-1} \vee \bigvee \left(\bigvee^{k_i} S^{n-1-k_i}\right). $$ Moreover, the inclusion of the outside boundary sphere is the fold map from $S^{n-1}$ to the first wedge of $(n-1)$-spheres.

Now finally, the complement $X$ of $k_1$ 1-spheres, etc, in the $n$-sphere, is the pushout $$ Y \leftarrow S^{n-1} \rightarrow D^n $$ where the left-hand map is the inclusion of the outside boundary sphere of radius $R$. Up to homotopy equivalence, this is the pushout of $$ \bigvee^{\sum k_i} S^{n-1} \vee \bigvee \left(\bigvee^{k_i} S^{n-1-k_i}\right) \leftarrow S^{n-1} \rightarrow D^{n-1}. $$ The left-hand map is the pinch map into the first wedge factor. Up to homotopy equivalence, gluing in this cell eliminates one sphere factor from the wedge; the resulting space is the one you were hoping for.

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[As noted in the comments, this answer missed the specification in the question that the spheres be unknotted and unlinked.]

Alexander duality, as noted in the comments, will compute the homology of the complement. But the fundamental group will not generally be free of the stated rank $k_{n-2}$, so you won't get a wedge of spheres for the homotopy type. This is already true for classical knots (eg a trefoil knot, or indeed any non-trivial knot) or link (eg the Hopf link).

If you disallow codimension 2 components (and so get a simply connected complement) it's still not true that the homotopy type is that of a wedge of spheres. The simplest example I can think of is a pair of 2-spheres linked in $S^5$ constructed as follows. Write $S^5 = S^2 \times B^3 \cup_{S^2 \times S^2} B^3 \times S^2$. Take the link comprised of $S^2 \times \{0\} \cup \{0\} \times S^2$; its complement is homotopically $S^2 \times S^2$, which is not a wedge of spheres.

I suggest a look at Rolfsen's book, Knots and Links, for the basics of the subject. If your primary interest is in the higher dimensional case, you will find a number of interesting constructions scattered in the book.

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  • $\begingroup$ Right, but I'm interested in an unlinked and an unknotted case (as stated at the end of my question), in which, I think, my suggestion should be correct. But I do not see how I can prove that they are homotopy equivalent.. $\endgroup$ – iou Jan 19 at 14:50
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    $\begingroup$ Sorry, didn't read carefully! I was preparing an answer to your actual question but others beat me to it. I would prove it inductively, using the splitting along (n-1)--spheres. $\endgroup$ – Danny Ruberman Jan 19 at 17:14

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