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If you have some embedding of a path connected topological space wedge of spheres $N$ into a compact simply connected smooth $n$ manifold $M$ (like a sphere for example), then

  1. is there some kind of analog to a tubular neighborhood for this sort of embedding? i.e. some $T$ that is a codimension 0 smooth submanifold of $M$ which deformation retracts to $N$, and

  2. what can easily be said about the (co)homology of such a space in relation to $N$ (and its ring structure)?

My main question is for wedges of spheres in a higher dimension sphere, so if the above seems too general then lets restrict to this. I think 1. is a silly question and I just don't have the right definition laying around.

For example, consider $S^1\vee S^1$ embedded in an $\mathbb{R}^3 \subset S^3$. Then a connected sum of two solid tori, call this $T$, satisfies my conditions. The cohomology of $S^1\vee S^1 $ has two basis elements in degree 1 with trivial product. The cohomology of $T$ has two generators in degree 1, two in degree 2, and one in degree 3 (there are two non-trivial products of a thing in degree one with a thing in degree 2). I forgot here that a solid torus has a boundary.

In a way, the cohomology of $T$ comes from adding dual elements in the right degrees. In general, for a wedge sum of spheres I think the example above sort of demonstrates whats going on. Does this seem right?

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  • $\begingroup$ Question 1 seems a little far fetched indeed, at least in full generality. $\endgroup$ – Olivier Bégassat Jul 22 '12 at 6:20
  • $\begingroup$ lets just stick with wedges of spheres then. $\endgroup$ – AnonymousCoward Jul 22 '12 at 6:24
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    $\begingroup$ This seems a little bit similar to an argument in Section 2 of a recent article I wrote with Tom Baird: arxiv.org/abs/1206.3341. There we construct a sort of "tubular neighborhood" for a normal crossings divisor in a non-singular algebraic set. But really the point is just that the normal crossings divisor is locally like a (nice) wedge of codimension 1 submanifolds. Anyway, the arguments could be of some use to you (maybe). $\endgroup$ – Dan Ramras Jul 22 '12 at 22:37

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