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Is the wedge sum of two $n$-manifolds homotopy equivalent to a manifold?

My guess is yes, at least under some suitable hypotheses. If they are both smooth, we can embed the wedge sum in some $\mathbb{R}^m$ for $m \geq n$ and find an open subset that deformation retracts onto the embedded space. I am less optimistic that such a wedge sum is in general homotopy equivalent to a closed manifold.

Is there somely pure algebraic-topological way to make sense of this question?

Paul Orland
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    Could you clarify: Do you require a closed manifold? The answer will depend on this. Also, are your manifolds topological or smooth? (This does not matter that much in the end.) – Moishe Kohan May 01 '22 at 21:55
  • I'm interested in topological manifolds, and a homotopy equivalence to any manifold (not necessarily closed). However, if there are interesting special cases I would love to hear about those too. – Paul Orland May 01 '22 at 22:17

1 Answers1

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Every topological manifold is homotopy-equivalent to a finite-dimensional locally-finite simplicial complex. (This should be in Hatcher's book, see also here and here.) Hence, the wedge sum of topological manifolds is also homotopy-equivalent to a finite-dimensional locally-finite simplicial complex. Lastly, embedding such a complex as a subcomplex in some ${\mathbb R}^N$ and taking an open regular neighborhood of the image yields a smooth open manifold homotopy-equivalent to the above wedge sum.

Of course, if you are insisting on closed manifolds, then the answer is quite different: if $M_1, M_2$ are closed connected $n$-dimensional manifolds ($n\ge 1$), then $H_n(M_1\vee M_2, {\mathbb Z}_2)\cong {\mathbb Z}_2^2$, which implies that $M_1\vee M_2$ cannot be homotopy-equivalent to a closed connected $n$-dimensional manifold. One sees that disconnected manifolds cannot be homotopy-equivalent to $M_1\vee M_2$ by looking at $H_0$. One similarly rules out closed manifolds of other dimensions.

Moishe Kohan
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