Finite dimensional CW complex $X$ other than wedge of two spheres so that $\tilde{H}(X) \approx \mathbb{Z} \oplus \mathbb{Z} $ I have been wanting to know distinct finite dimensional CW complexes $X$ upto homotopy equivalence with the homology $$\tilde{H}(X):=\oplus_{i \in \mathbb{N}}\tilde{H}_i(X)\approx \mathbb{Z} \oplus \mathbb{Z}$$
So far, using the Mayer-Vietoris sequence, I know that a wedge sum of two spheres $\mathbb{S}^k \vee \mathbb{S}^n$ have such a property. How can I find or disprove whether or not there exists any other CW complexes with such a property ?
 A: Assuming you mean the reduced homology, the complex projective plane $\mathbb{C}P^2$ is another space with $\tilde{H}_*(\mathbb{C}P^2) \cong \mathbb{Z} \oplus \mathbb{Z}$.  (To see that it is not a wedge of spheres, consider the cup product in cohomology.)
A: Generalising JHF's answer, take any $n\geq2$ and any nontrivial element $\varphi\in\pi_{n+k}S^n$ where $k\geq1$.

Then $X=S^n\cup_\varphi e^{n+k+1}$ has the correct homology and is not homotopy equivalent to a wedge of spheres.

The constraint $k\geq1$ is to ensure that $\varphi$ induces the trivial map in homology, so that a quick inspection of the long exact sequence of the pair $(X,S^n)$ will verify that $X$ has the correct homology.
The constraint that $\varphi$ is nontrivial is to ensure that $X$ is not homotopy equivalent to a wedge of spheres. For the universal property of the pushout shows that null-homotopies of $\varphi:S^{n+k}\rightarrow S^n$ (that is, to extensions of this map over $D^{n+k+1}$) correspond to retractions $X\rightarrow S^n$. Suppose that $X$ is homotopy equivalent to $S^n\vee S^{n+k+1}$. Then the composite $X\simeq S^n\vee S^{n+k+1}\rightarrow S^n$ gives a map $r:X\rightarrow S^n$, which we can assume to restrict to $S^n$ as a degree $1$ map (looking at homology shows that $r|_{S^n}$ is a map of degree either $+1$ or $-1$. In the latter case replace $r$ with its composition with another degree $-1$ map). Thus $r|_{S^n}\simeq id_{S^n}$, and since $S^n\subseteq X$ has the homotopy extension property, we can extend the previous homotopy to $r\simeq r'$, where $r':X\rightarrow S^n$ satisfies $r'|_{S^n}=id_{S^n}$. In particular $\varphi\simeq\ast$, which contradicts the basic assumption about $\varphi$.
Finally, the constraint $n\geq2$ is to ensure that the construction is not trivial. It is well known that if $n\geq2$, then $\pi_{n+k}S^n$ contains nontrivial elements for infinitely many $k$. In fact, when $n=2,3,4,5$, then $\pi_{n+k}S^n$ contains nontrivial elements for all $k\geq0$.
A: There are things called homology spheres (see https://en.wikipedia.org/wiki/Homology_sphere). For example, the Poincare homology sphere has the same homology as a $3$ sphere (but not the same homotopy type), so one way to produce more examples is to take wedge sums of homology spheres.
