It's pretty clear that for a simplicial complex $\Delta$, shellability of the complex implies that it is homotopy equivalent to a wedge of spheres. However, does the converse hold? That is, does $\Delta$ being homotopy equivalent to a wedge of spheres imply shellability? Thanks!



Does homotopy equivalence to a wedge of spheres imply shellability?
No. There are unshellable n-spheres for all n≥3:


*Lickorish, W. B. R.*, [**Unshellable triangulations of spheres**](http://dx.doi.org/10.1016/S0195-6698(13)80103-5), Eur. J. Comb. 12, No. 6, 527-530 (1991). [ZBL0746.57007](https://zbmath.org/?q=an:0746.57007).


