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The given homeomorphisms and homotopy invariance of homology givebecause $n\ge 2$, while $M\simeq S^{n-1}$ is path connected and $L$ is the disjoint union of two copies of $S^{n-1}$. HenceThe degree-one and degree-zero part of the Mayer–Vietoris sequence is thereforeTaking the rank of an abelian group throughout this short exact sequence givesso $\mathrm{rank}{H}_0(N)=2$. Since the zeroth homology group is freely generated by the path components,

The inductive definition of ordinal addition isfor a limit ordinal $\lambda$.

The synthetic definition takes the order type of the disjoint union of a copy of $\alpha$ followed by a copy of $\beta$: every point of the first copy precedes every point of the second, and each copy retains its original total order.

Fix $\alpha$ and use transfinite induction on $\beta$. The synthetic sum with the empty order is $\alpha$. Appending a greatest element produces the successor ordinal rule. At a limit ordinal $\lambda$, the second copy is the union of its initial segments of types $\beta <\lambda$, so the whole order has type
${\sup }_{\beta <\lambda }(\alpha +\beta )$. Thus the synthetic operation satisfies the inductive recursion; uniqueness in transfinite recursion proves the definitions equivalent.

Part (b) proves one implication. Conversely, assume in ZF thatfor every infinite cardinal $\kappa$. Let $X$ be any set and let $\gamma =h(X)$ be its Hartogs ordinal. Finite $X$ is already well-orderable, so suppose $X$ is infinite and form the disjoint unionBy hypothesis, $A\times A$ is bijective with $A$. Since $\gamma \times X$ injects into $A\times A$, we obtain

Apply part (c), with $K=\gamma$ and $L=X$. It gives either an injection $\gamma \to X$ or a surjection $s:\gamma \to X$. The first alternative contradicts Hartogs theorem. In the second, every fiber $s^{-1}(\{x\})$ is a nonempty set of ordinals and therefore has a least member. The mapis an injection of $X$ into the ordinal $\gamma$ and pulls its well-order back to $X$. Thus every set is well-orderable, so the well-ordering theorem gives the axiom of choice. This proves the Tarski cardinal-square theorem.