A question about the category of sets with endomorphisms

Question

I'm having a hard time to figure out a problem my professor assigned in an introductory course in category theory. I'm a philosophy student and I haven't much background in math. It's about the category of sets with endomorphisms, which is defined like this (I couldn't find this example of category in any book, but is pretty straightforward):

The category of sets with endomorphisms is the category whose objects are sets $A, B ...$ equipped each one with one endomorphism $\alpha, \beta ...$. A morphism $$f:(A,\alpha)\rightarrow(B,\beta)$$ is a function from the set $A$ to set $B$ that respects the endomorphisms, i.e, $$f\circ\alpha=\beta\circ f$$ Now, the problem is: in this category, consider the objects $(\mathbb Z, \alpha)$ and $(\mathbb Z,\beta)$ where $\alpha(n)=2n$ and $\beta(n)=3n$. Are this two objects isomorphic?

And when would two such objects (with the underlying set $\mathbb Z$ but two different endomorphisms) be isomorphic?

I ask this more general question as well because in fact I can't even find a morphism between this two objects, much less one that has an inverse. The only morphism I could think of, that respects the endomorphisms, is the one that sends every number to zero, but this one can't have an inverse. So i'm tempted to say that the answer to the problem is negative, but I can't really prove it. What am I missing in this problem?

(Also, this is my first question here and I would accept any suggestion on how to improve it)

Answer 1

Let me try to give some insight into how to think about objects of this category. First, forgetting about the endomorphisms, what is a set? It's just some collection of points. So, you should envision this as just a big bag of points with no structure relating them.

Now, if you have such a set $A$, what additional structure does an endomorphism $\alpha:A\to A$ give you? Well, for each element $a\in A$, it links $a$ to some other element $\alpha(a)$. You could visualize this by imagining that you draw an arrow from $a$ to $\alpha(a)$, for each point in your set. So, instead of just a bag of points, you now have a diagram of points where there are some arrows between them, with exactly one arrow starting at each point. (More formally, a mathematician would call this a special type of "directed graph".)

Let's look at some simple examples. First, consider $A=\mathbb{N}$ with $\alpha(n)=n+1$. This looks like an infinite "ray" of points going off the the right, with an arrow from each point to the next one. Another example is $B=\mathbb{N}$ with $\beta(n)=n+2$. This has the same set of points, but now the arrows are different so they form two separate "rays". One ray consists of all the even numbers, and the other ray consists of all the odd numbers. We can thus see that $(A,\alpha)$ and $(B,\beta)$ are not isomorphic, since $(A,\alpha)$ has just a single connected ray while $(B,\beta)$ consists of two separate rays which are not connected to each other at all via the map $\beta$. (This isn't a completely rigorous argument, but it can be made into one.) Another example is $C=\mathbb{Z}$ with $\gamma(n)=n+1$. This one has just one connected piece like $(A,\alpha)$, but now instead of a ray it is a line that continues infinitely both forward and backward.

Your examples are quite a bit more complicated, but can be analyzed in a similar spirit. So, here are some things I encourage you to contemplate on your own. For $(\mathbb{Z},\alpha)$ with $\alpha(n)=2n$, what "connected pieces" can you break the picture into? What shape are the pieces? Are they rays? Lines? Something else? And, can you do the same analysis for $(\mathbb{Z},\beta)$ with $\beta(n)=3n$? Does it have the same number of pieces with the same shape (so it would be isomorphic), or not?

Brief answers to these questions are hidden below.

For $(\mathbb{Z},\alpha)$, there are infinitely many pieces. Namely, for any odd integer $n$, the set of integers of the form $2^kn$ for $k\geq 0$ forms a "ray", starting at $2^0n=n$ and then with $\alpha$ taking each $2^kn$ to the next point $2^{k+1}n$. Besides these rays, there is one other piece, consisting of just $0$. This is a "loop" with one point, since $\alpha$ maps $0$ to itself. So, to sum up $(\mathbb{Z},\alpha)$ consists of (countably) infinitely many rays together one one single point loop.

What about $(\mathbb{Z},\beta)$? It's almost the same! We again have infinitely many "rays", except now they start at integers that are not multiples of $3$, rather than at odd integers. And besides these rays, we have a one point loop at $0$.

So, since $(\mathbb{Z},\alpha)$ and $(\mathbb{Z},\beta)$ each consist of countably infinitely many rays and one single point loop, they should be isomorphic. I leave it to you to try to prove this more formally.