I am working through Conceptual Mathematics (Lawvere/Schanuel) and am stuck on the problem in Session 11/ problem #3 (on pg 159 in my edition).
The problem is:
Given:
$\mathbb Z$ is the set of all integers and $\mathbb Z^{\circlearrowleft \alpha}$ and $\mathbb Z^{\circlearrowleft \beta}$ are the endomaps which add 2 and 3 to their inputs. That is
${\alpha}$(x) = x + 2
${\beta}$(x) = x + 3
Prove that in the category of sets-with-endomaps there is no isomorphism between $\mathbb Z^{ \circlearrowleft \alpha}$ and $\mathbb Z^{\circlearrowleft \beta}$.
My attempt at a proof
In the category of sets with endomaps any map $f$ $ | f: \mathbb Z^{\circlearrowleft \alpha} \rightarrow \mathbb Z^{\circlearrowleft \beta}$ must obey the law
$$f \circ \alpha = \beta \circ f$$
So, $$f \circ \alpha(x) = \beta \circ f(x)$$ $$f(\alpha(x)) = \beta(f(x))$$ $$f(x+2) = f(x)+3$$
I need to prove that the last line leads to a contradiction, but at this point I get stuck. I couldn't figure out how to apply proof by induction, and ran into several other dead ends. Any help would be most gratefully appreciated ! thanks /chris
Suppose that $f: \mathbb Z^{\circlearrowleft \alpha} \rightarrow \mathbb Z^{\circlearrowleft \beta}$ is a morphism. Then $f(0) = a$ and $f(1)=b$ for some $a, b \in \mathbb{Z}$. Show by induction that $f(2k) = a+3k$ and $f(2k+1)= b + 3k$ for $k \in \mathbb{Z}$. Can you see that $f$ can't be surjective? To do so you need to explain why there must be a number $c \in \mathbb{Z}$ such that $a - c$ and $b-c$ are not divisible by $3$.