The integral Kronecker map need not be an isomorphism

By [F1], the right term of [F2] is $\mathrm{Hom}(0,\mathbb{Z})=0$ and its left term is ${\mathrm{Ext}}^1(\mathbb{Z}\mathrm{/}2,\mathbb{Z})$. The latter is computed from the exact free resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$. Hom into $\mathbb{Z}$ gives multiplication by two from degree zero to degree one, with zero next differential. Thus its first cohomology is $\mathbb{Z}\mathrm{/}2$. The homomorphism $\psi :\mathbb{Z}\to \mathbb{Z}$, $\psi (1)=1$, represents its nonzero class: it cannot be a boundary, since precomposition by multiplication by two always has even value at $1$.

Transport that presentation class by the canonical comparison of [F3] to an element $e$ of the Ext term in [F2], and put $\alpha =\iota e$. Injectivity of $\iota$ gives $\alpha \mathrm{\ne }0$. Exactness gives $\beta \alpha =0$, and since the entire right term is zero, $\iota$ is also onto. Consequently $H^2(X;\mathbb{Z})\cong \mathbb{Z}\mathrm{/}2$ and evaluation is the zero map from this nonzero group. This explicit nonzero presentation class and its injective image witness failure of injectivity, and hence of being an isomorphism.

Equivalently every integral singular two-cycle is a boundary because $H_2=0$. Any degree-two cocycle vanishes on each such cycle by its cocycle equation, including a representative of $\alpha$. Therefore all pairings with $\alpha$ vanish even though its cohomology class is nonzero. Also $2\alpha =0$, while the generator value one in the presentation is nonzero modulo two. The space is nonempty and finite dimensional; failure is neither a negative-degree convention nor an infinite-rank phenomenon. AC is inherited from [F2] and its comparison supplier, not from the explicit two-term integer calculation.