Construct a CW Complex $X$ with the following homology groups (in coefficiants $\mathbb{Z})$: $H_0(X)=\mathbb{Z}, H_1(X)=\mathbb{Z}\oplus\mathbb{Z_2}, H_2(X)=\mathbb{Z_3}, H_3(X)=\mathbb{Z}, \; $ and $ \; H_n(X)=0 \;$ if $\; n\ge 4$.
Is my construction a correct way to think about these kinds of problems? Any comments/feedback & other methods are welcome.
Let $X^0=e_0$ and let $X^1$ have two 2-cells $e_1^1, e_2^1$ to obtain the wedge of 2 circles. Attach a 2-cell $e_1^2$ to $X^1$ such that $\partial_2(e_1^2)=2e_2^1$. At this point, we have $H_0(X)=\mathbb{Z}$ and $H_1(X)=\mathbb{Z}\oplus \mathbb{Z_2}$. In particular, $H_2(X)=0\; $ since the boundary map $\partial_2$ is injective.
To amend this, we can attach a second 2-cell $e_2^2$ to $X^0$; this necessarily gives $\partial_2(e_2^2)=0$. Thus $\textrm{ker}(\partial_2)\cong\mathbb{Z}$. Attach a 3-cell $e_1^3$ to $X^2$ via $\partial_3(e_1^3)=3e_2^2$ and so $H_2(X)=\mathbb{Z_3}$. Here $H_3(X)=0$.
Similarly, attach a second 3-cell $e_2^3$ to $X^0$ such that $\partial_3(e_2^3)=0$. This gives $H_3(X)\cong\mathbb{Z}.\;$ Let $X=X^n\;$ for $\;n\ge 4$ and so $H_n(X)=0 \;$ for $\; n\ge 4$ and we are done.