Consider a poset-subcategory of the category of CW-complexes, the morphisms of which are only inclusions of deformation retracts (since we restricted ourselves to CW-complexes, this is [equivalent](https://mathoverflow.net/questions/54299/about-deformation-retract) to "inclusion that is a homotopy equivalence"). The minimal elements in it will be called homotopy incompressible. That is, the space is homotopically incompressible iff it does not have its proper deformation retract.

1) It is clear that among the 1-dimensional CW-complexes these are exactly bouquets of circles, and all the other one-dimensional ones are deformation retracted on them.

2) $S^n$ are homotopy incompressible since [no subspace $\mathbb{R}^n$ is equivalent to $S^n$](https://math.stackexchange.com/questions/3649728/can-a-subset-of-mathbbrn-be-homotopy-equivalent-to-sn?rq=1). So in each dimension there is a homotopy incompressible space.

3) In general, [closed manifolds are homotopically incompressible](https://math.stackexchange.com/questions/258367/is-it-possible-for-a-closed-manifold-to-deformation-retract-onto-a-proper-subset?noredirect=1&lq=1) (in particular, spheres with handles and films are incompressible).

Questions:

4) Is it possible to classify homotopically incompressible 2-dimensional CW-complexes?

5) Is it true that each CW-complex has a homotopically incompressible deformation retract (i.e., do the minimal elements in our order form a barrier from below / are there any chains going as many downward as desired)?
 
Due to (3), it is interesting - is it true that every connected n-dimensional CW-complex is homotopy equivalent to a bunch of n-dimensional closed manifolds? What is the criterion for incompressibility for non-closed manifolds? Are homotopy incompressible spaces closed with respect to a bouquet, product ..

**Update 1**: of course, the homotopy types of CW-complexes are not limited to bouquets of closed n-manifolds since the latter have a restriction on n-dimensional homology (first comment).