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A simply connected space is path-connected and has trivial fundamental group, equivalently every based loop is null-homotopic. The Whitehead theorem says that a map of CW complexes inducing isomorphisms on all homotopy groups is a homotopy equivalence. Its homological Whitehead theorem form says that a map between simply connected CW complexes, in particular finite ones, is a homotopy equivalence if it induces isomorphisms on integral homology. The latter implication follows from the relative Hurewicz theorem and the ordinary Whitehead theorem; homotopy equivalences necessarily induce homology isomorphisms.

For the attached-cell space $X$, the cellular homology chain complex in positive degrees isDualizing gives the cellular cohomology complex $\mathbb{Z}\stackrel{0}{\to }\mathbb{Z}\stackrel{t\mapsto (2t,3t)}{\to }{\mathbb{Z}}^2$. The last map is injective and its image is primitive because $\gcd (2,3)=1$. Its cokernel is $\mathbb{Z}$, explicitly via $(a,b)\mapsto 3a-2b$. HenceThe van Kampen theorem gives ${\pi }_1(X)=⟨a\mid a^2=1,a^3=1⟩=1$. Moreover $H_2(X)=\ker (23)$ is generated by $(3,-2)$. Since $X$ is simply connected, the Hurewicz theorem gives ${\pi }_2(X)\cong H_2(X)$. Choose a map $S^2\to X$ representing that generator. It induces an isomorphism on $H_2$, and on $H_0$, while all other reduced homology groups vanish on both sides. The homological Whitehead theorem proves $\boxed{X\simeq S^2}$. This is a special case of coprime two-cell attachments to a circle.

It is not homeomorphic to the sphere. Indeed, if it were, its embedded one-skeleton would be a Jordan circle in $S^2$, with the two open cell interiors as its complementary regions. The Jordan curve theorem and its Schoenflies form identify the closures of those regions with discs. A characteristic map, homeomorphic on the interior of such a disc, has local degree $+1$ or $-1$ at an interior point; the boundary homomorphism in relative homology then makes its attaching map on the circle have degree $+1$ or $-1$. This contradicts degrees $2$ and $3$. This argument does not require choosing special covering representatives of the attaching maps.

Finally, two compact examples with the same additive cohomology areBoth have $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^2$, $H^2=\mathbb{Z}$ and zero higher cohomology, by their cellular chain complexes. But their fundamental groups are respectively ${\mathbb{Z}}^2$ and the nonabelian free group on two generators, so they cannot be homotopy equivalent. This illustrates that cohomology groups do not determine homotopy type.