Wedge sum of manifolds is homotopy equivalent to manifold? 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?
 A: 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.
