Suppose $X$ is a finite $d$-dimensional simplicial complex which is homotopy equivalent to a wedge of at least two $d$-spheres. Does $X$ contain a subcomplex which is homotopy equivalent to a single $d$-sphere?



Subcomplexes with homotopy type of a sphere in complexes with homotopy type of a wedge of spheres
EDIT: Wow, I worked a lot harder than necessary. I've added a simplified version of this proof. The original is at the bottom.


Let $A = S^1$, with fundamental group the free group $F$ on a generator $x$. Consider the elements
$$
\begin{align\*}
a &= x^6\\
b &= x^{10}\\
c &= x^{15}
\end{align\*}
$$
in $F$, and use them as attaching maps: attach three 2-dimensional cells to $A$ to construct a space $B$.


Here are some properties of $B$.


* It is simply-connected. The fundamental group of $B$ is
$$
\langle x \mid x^6, x^{10}, x^{15}\rangle
$$
and this group is trivial.
* It has homology $\Bbb Z^2$ in degree $2$, $\Bbb Z$ in degree 0, and 0 elsewhere. You can see this from the cellular chain complex $0 \to \Bbb Z^3 \to \Bbb Z \to \Bbb Z \to 0$ for computing $H\_\* B$, using the fact that $H\_1 B = [\pi\_1 B]\_{ab} = 0$.
* As a result, $\pi\_2(B) = \Bbb Z^2$ by the Hurewicz theorem and there is a map $S^2 \vee S^2 \to B$ inducing an isomorphism on $H\_\*$. By the Whitehead theorem, this means that $B$ is homotopy equivalent to $S^2 \vee S^2$.


I claim that no proper subcomplex of $B$ is homotopy equivalent to a $2$-sphere because any proper subcomplex that has some 2-dimensional cells is not simply-connected.


The only possibility is for such a subcomplex to take the $1$-cell and some assortment of the $2$-cells. However, taking any two of the 2-cells doesn't give you something simply-connected, because you only get these possible fundamental groups:


* $\langle x \mid x^{10}, x^{15}\rangle \cong \Bbb Z/5$
* $\langle x \mid x^6, x^{15}\rangle \cong \Bbb Z/3$
* $\langle x \mid x^6, x^{10}\rangle \cong \Bbb Z/2$


(Using fewer than two of the $2$-cells gives you an even larger fundamental group.)




---


The original example had $A = S^1 \vee S^1$, with fundamental group the free group $F$ on generators $x$ and $y$, and used
$$
\begin{align\*}
a &= x^2\\
b &= y^3\\
c &= (xy)^5\\
d &= xyx^{-1}y^{-1}
\end{align\*}
$$
as attaching maps for four 2-dimensional cells.


