Let $K$ be a finite $d$-dimensional simplicial complex which is homotopy equivalent to a wedge of $k$ $d$-spheres. Does $K$ necessarily contain a $d$-sphere as a subcomplex?
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1$\begingroup$ I'll note that it's true for d=1, since the fundamental group of K is surjected onto by the fundamental group of the 1-skeleton of K, and hence that 1-skeleton contains a simple closed loop. $\endgroup$– Dan RamrasCommented Nov 3, 2016 at 18:12
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It is false for $d\geq 2$. For an example with $d=2$, start with a circle, and attach two 2-discs to it via (for example) a map of degree 4 and a map of degree 5. If you like to think in terms of group presentations, this is the presentation 2-complex for the presentation $\langle a: a^4=a^5=1 \rangle$ of the trivial group. This 2-complex is homotopy equivalent to a single copy of the 2-sphere, but does not contain any subcomplex homeomorphic to the 2-sphere.
Describing an explicit triangulation of a space homeomorphic to this is a bit more complicated, but it can be done. Triangulate the circle using four edges, and triangulate the two discs so that their boundary circles have 16 and 20 edges respectively, and so that there are no `short routes' in the interior of the disc between distinct vertices on the boundaries of the disc. Given any such triangulations, you can glue the discs on to the circle and get a simplicial complex homeomorphic to the CW-complex mentioned above. If you want to find a subcomplex that is homeomorphic to the 2-sphere, it must contain one of the 2-simplices. But once it contains one of the 2-simplices, it has to contain every 2-simplex in the triangulation of one of the two 2-discs, in which case it contains an edge of valency 4 or 5, giving a contradiction.
Suspensions of this example (suitably triangulated) will work in higher dimensions by a similar argument.
