Let $X$ be an $(n-1)$-connected CW complex of dimension $n$ and $\{\varphi_i : \mathbb{S}^n \to X \mid i \in I\}$ a generating set of $\pi_n(X)$. If $\dot{X}$ denotes the complex obtained from $X$ by gluing $(n+1)$-cells via the $\varphi_i$'s, is the image of $\pi_{n+1}(X)$ in $\pi_{n+1}(\dot{X})$ trivial?
In other words, does the fact that $X$ is $n$-dimensional imply that homotopically non-trivial spheres $\mathbb{S}^{n+1} \to X$ come from spheres $\mathbb{S}^n \to X$?
Best Answer
If $X$ is $(n-1)$-connected and $n$-dimensional, then $X$ is homotopy equivalent to a wedge of spheres.
If $n=1$ this is true because any path-connected CW complex $X$ is homotopy equivalent to a CW complex with exactly one $0$-cell. In this case $X$ is a graph. A subcomplex $Y$ which is a maximal tree can be found. Then $Y\simeq \ast$. Moreover the subcomplex inclusion $Y\hookrightarrow X$ is a cofibration. It follows that the quotient $X\rightarrow X/Y$ is a homotopy equivalence, and the standard cell-structure on $X/Y$ has exactly one zero-cell.
If $n\geq2$ then $\pi_nX$ is necessarily free abelian. Choose a minimal set of generators out of the $\varphi_i$ to get a map $\theta:\bigvee S^n\rightarrow X$ which induces an isomorphism on $\pi_n$, and so on $H_n$ by the Hurewicz Theorem. Since the cellular complex of $X$ contains no generators in higher dimensions $\theta$ is in fact an isomorphism on $H_*$. Thus by the Homological Whitehead Theorem $\vee \varphi_i$ is a homotopy equivalence.
Edit regarding comments: We identify $X\simeq\bigvee S^n$ as above. Then up to homotopy $\Phi:\bigvee\varphi_i:\bigvee_{I} S^n\rightarrow \bigvee S^n=X$ has a section. It follows that the map $X\rightarrow C_{\Phi}=\dot X$ is nullhomotopic. Since all the $\varphi_i$ are suspensions, the space $\dot X$ is a suspension. It follows that $\dot X$ is homotopy equivalent to a wedge of $S^{n+1}$, one for each reduntant generator amongst the $\varphi_i$.