Let $A$ and $B$ be abelian groups. Say that a map $f: A \to B$ is a quasi-homomorphism if there exists a finite $D \subseteq B$ such that $$\forall a_1, a_2 \in A: f(a_1 + a_2) - f(a_1) - f(a_2) \in D$$

We wish to show that if $A$, $B$ and $C$ are abelian groups and $f: A \to B$ and $g: B \to C$ are quasi-homomorphisms, then $g \circ f: A \to C$ is a quasi-homomorphism. So take a finite $D \subseteq B$ such that $$\forall a_1, a_2 \in A: f(a_1 + a_2) - f(a_1) - f(a_2) \in D$$ and a finite $E \subseteq C$ such that $$\forall b_1, b_2 \in B: g(b_1 + b_2) - g(b_1) - g(b_2) \in E$$ Define $$F = \{g(d) + e_1 + e_2 \mid d \in D, e_1, e_2 \in E\} \subseteq C$$ As $D$ and $E$ are finite, $F$ is also finite. Now take any $a_1, a_2 \in A$. We wish to show that $$g(f(a_1 + a_2)) - g(f(a_1)) - g(f(a_2)) \in F$$ Now, set $$d = f(a_1 + a_2) - f(a_1) - f(a_2) \in D$$ Set $$e_1 = g(d + f(a_1) + f(a_2)) - g(d) - g(f(a_1) + f(a_2)) \in E$$ Set $$e_2 = g(f(a_1) + f(a_2)) - g(f(a_1)) + g(f(a_2)) \in E$$ Then we have $$g(f(a_1 + a_2)) - g(f(a_1)) - g(f(a_2)) = g(d + f(a_1) + f(a_2)) - g(f(a_1)) - g(f(a_2)) = e_1 + g(d) + g(f(a_1) + f(a_2)) - g(f(a_1)) - g(f(a_2)) = e_1 + g(d) + e_2 \in F$$ Thus, $g \circ f$ is a quasi-homomorphism.

In particular, we can form a category $\mathsf{QuasiAb}$ whose objects are abelian groups and whose morphisms are quasi-homomorphisms.

Is this proof correct? Can you give any reference on $\mathsf{QuasiAb}$?

  • 1
    $\begingroup$ From the solution-verification tag wiki: "A question with this tag should include an explanation for why the argument presented is not convincing enough." $\endgroup$
    – Shaun
    Apr 14 at 12:19
  • $\begingroup$ Hmm... maybe a reference-request tag could be added, so that the question doesn't have to be closed just because a rule that the OP was unaware of wasn't followed. To me, the question is decent because this is not exactly a well-known concept. $\endgroup$
    – user43208
    Apr 14 at 13:37

1 Answer 1


I mean, sure, the proof is correct.

The late Stephen Schanuel proposed a remarkable construction of the real numbers, calling them the "Eudoxus real numbers", as the set of all "quasi-homomorphisms" [or "almost linear maps", as he calls them] $\mathbb{Z} \to \mathbb{Z}$, modulo the equivalence relation $\sim$ where $f \sim g$ if the set of differences $f(m) - g(m)$ is a finite set. You can check that the sum of two almost linear functions $\mathbb{Z} \to \mathbb{Z}$ is almost linear, and since (as you showed) the composite of two almost linear functions is almost linear, you get a ring structure on the set of almost linear functions, and this ring structure also respects the equivalence relation $\sim$, so that you get a commutative ring structure on the Eudoxus reals.

It can be shown that this commutative ring is a field. From the nLab: Define a Eudoxus real $[f]$ to be positive if it is represented by an $f$ such that for any $C$ there are infinitely many $m\in \mathbb{N}$ such that $f(m)>C$. It is easy to see this condition would then be satisfied by any representative function. As usual, define $[f]<[g]$ to mean $[g−f]$ is positive. Then the Eudoxus reals forms an ordered field, and in fact it is a complete ordered field.

Mathematics StackExchange user Rob Arthan gave a nice account of this construction here, and you can find some more information at the nLab, which I cited above.

  • 3
    $\begingroup$ Neat example, I hadn't heard of it! To demystify it a little bit, there's an explicit correspondence with the real numbers where $r\in\mathbb R$ maps to (the equivalence class of) the function $n\mapsto \lfloor rn\rfloor$. $\endgroup$
    – Carmeister
    Apr 15 at 4:45
  • 1
    $\begingroup$ Thanks, yes, I should have mentioned that. $\endgroup$
    – user43208
    Apr 15 at 9:52

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