**Edited to fix the example, as per Zack's suggestion.**

Edit 2: So it turns out that when I think 'manifold' I tend to assume the nicest possible object. As I believe is standard, I would like to assume that all manifolds are 2nd countable and Hausdorff. Furthermore, let's say that our manifolds are connected and closed.

The Whitney embedding theorem states that any smooth $n$-manifold may be smoothly embedded into $\mathbb{R}^{2n}$. If we consider embeddings into more general $k$-dimensional manifolds, is it possible find a '$n$-universal' manifold of dimension less than $2n$?

For example, a non-orientable 2-manifold cannot be embedded into $\mathbb{R}^3$, demonstrating the sharpness of the Whitney embedding theorem.

However, there are 3-manifolds into which we can embed any surface, such as $M = \mathbb{RP}^3 \sharp \mathbb{RP}^3$. Indeed, by the classification of surfaces we know that any surface may be decomposed as a connected sum of copies of $\mathbb{RP}^2$ and tori. In fact, by the monoid structure of closed surfaces under connected sums we may take this sum to have at most 2 copies of the projective plane. Now, embed 2 disjoint copies of the projective plane into $M$ and arbitrarily many copies of the torus. Taking the connected sum of these we see that any closed surface is embeddable into $M$.

Can we do something similar in higher dimensions?

cando this sort of has to be yes. 3. Certainly we can compare $n$-universal manifolds if their dimensions are different, but a priori there might be two $n$-universal $N$-manifolds and neither covers the other. Or perhaps it's provable that onemustcover the other; that'd be a neat result. – Aaron Mazel-Gee Mar 6 '11 at 17:35