Embeddability of connected sum of non-embeddable surfaces

Question

Let $X$ be a surface which can not be embedded into $\Bbb R^n$. Let $X \# X $ denotes the connected sum of two copies of $X$. Then is it true that the connected sum $ X \# X $ is also not embeddable into $\Bbb R^n$?

Answer 1

A closed orientable surface embeds into $\Bbb R^3$; a closed non-orientable surface does not (codimension 1 connected closed manifolds in $\Bbb R^n$ separate it into two pieces; one is compact, with boundary the closed manifold; the boundary of an oriented manifold is oriented). Because $X \# Y$ is oriented iff $X$ and $Y$ are, $X \# X$ embeds into $\Bbb R^3$ iff $X$ does. Since every compact surface embeds into $\Bbb R^4$ (Whitney's embedding theorem), the only question remains: what about surfaces with boundary?

We have the best possible result here: every compact surface with boundary embeds into $\Bbb R^3$. Think about the standard immersion of the Klein bottle in $\Bbb R^3$: it intersects itself precisely in a circle. Delete a disc from the Klein bottle that contains (one of the) circles of self-intersection. Then this embeds. You can see a picture of what I mean here on page 8.

Now more generally, every compact surface is a closed surface with discs deleted (it doesn't matter where you delete them; it does matter how many). It suffices to show that every compact surface with a single disc deleted embeds into $\Bbb R^3$. I will not give a detailed proof, but here is a sketch. You can immerse every closed non-orientable surface in $\Bbb R^3$ such that it intersects itself in a disjoint union of $2n$ circles (these circles 'pair off', since each self-intersecting circle in $\Bbb R^3$ is the image of two circles in the manifold). You can isotope this embedding so that the first $n$ (the first of each pair) are all arbitrarily close to each other; delete a disc containing these $n$. Then the immersion is now actually an embedding; there is no self-intersection, as desired.

So again, $X \# X$ emeds iff $X$ does.

Now, if you generalize your question to $m$-manifolds instead of surfaces, this is no longer true. For $m=3$, no lens space embeds into $S^4$, but $L(3,1) \# L(3,1)$ does. This is a special case of Theorem 2.16 in this survey by Ryan Budney on embedding 3-folds into $S^4$; references are given alongside the theorem.