By $Arrow(Set)$ I mean the category whose objects are the arrows of the Set category and whose arrows from the object $f : A \rightarrow B$ to the object $f' : A' \rightarrow B'$ are the pairs of functions $(a,b)$ such that $f' \circ a = b \circ f$
What are the final objects of the $Arrow(Set)$ category
category-theory
Related Solutions
There are already many excellent answers, but I want to add another perspective, already partly found in other answers, but I hope distinct enough to stand on its own.
I like to explain by analogy. Consider the question, "What is a vector?"
What is a vector?
Well, you might get you any of the following informal definitions as a response:
(a) a list of numbers, (b) a quantity with magnitude and direction, (c) a quantity that transforms like a vector under a change of coordinates.
(I think I might know some physicists who would take issue with me calling (c) informal, but oh well).
And you might then ask, well ok, but what's a formal definition of a vector?
Let's think of some examples of vectors. The elements of $\Bbb{R}^3$, the elements of $\Bbb{R}[x]$, continuous functions from $X$ to $\Bbb{R}$, where $X$ is a topological space. These seem like fairly different objects, but the common factor here is that a vector is simply an element of a set $V$ with a specified vector space structure. I.e., the formal definition of vector is simply an element of a vector space.
Why is this useful as a definition? Well, all of the properties of vectors are already encoded in the definition of vector space. So if I tell you that $v,w$ are vectors in $V$, and $r\in\Bbb{R}$ is a scalar, then you know that $v+w$ is also a vector, and that $rv$ is a vector, and that $r(v+w)=rv+rw$. All the properties of a vector that we might find interesting are encoded in the vector space axioms.
Note also that part of this means that it's meaningless to say $v$ is a vector on its own. It's only meaningful to say that $v$ is a vector of some vector space $V$. This is good, because as an element $v$ might belong to many different vector spaces with different structures, but depending on the ambient vector space structure, $v$ might behave completely differently.
Bringing it back to arrows
Similarly, if I say $f:X\to Y$ is an arrow of a category $\mathcal{C}$. The rigorous definition of arrow here is simply that $f$ belongs to the collection of arrows $\operatorname{Arr}(\mathcal{C})$, and that the domain of $f$ is $X$ and the codomain of $f$ is $Y$. All of the other interesting properties of arrows (for example that I could compose $f:X\to Y$ with an arrow $g:Y\to Z$ to get an arrow $g\circ f:X\to Z$) are already encoded in the axioms of the category $\mathcal{C}$, and there's no need to say anything further to define arrows.
Edit:
Since comments are not permanent, I just want to edit in the link from Ethan Bolker's comment, to an excellent answer with a similar viewpoint to this one in reply to a similar (in spirit) question about "what actually is a polynomial?" The second paragraph in particular really captures what I wanted to say in my answer, (paraphrasing Ethan's answer) what really matters isn't what something actually is, but rather how it behaves.
Let $g : (\mathbb N, s,0) \to (A,\alpha, a)$ be an arrow. $g(0) = a = \alpha^0(a)$ holds by assumption. Suppose $g(n) = \alpha ^n(a)$ for some $n\in\mathbb N$. Then $g(n+1) = gs(n) = \alpha g(n) = \alpha (\alpha ^n(a)) = \alpha ^{n+1}(a)$. So we have an initial object in the category.
Best Answer
I have tried to give you a way of seeing what the answer should be (without telling you precisely).
To find a terminal (final) object we want a map $t: S\to T$ such that for all maps $f: A\to B$ there exist unique maps $a :A \to S$ and $b: B\to T$ such that $ta=bf$. For convenience lets write such a morphism as $(a,b):f\to t$. Suppose such a $t$ exists. Now for any two maps $u : W\to S$ and $v : W\to S$ we obtain morphisms $(u,tu): 1_W \to t$ and $(v,tv) : 1_W \to t$. What can we conclude about $(u,tu)$ and $(v,tv)$ and hence about $u$ and $v$? On the other hand for any set $C$ there must be a morphism from $1_C$ to $t$ and hence a map from $C$ to $S$. What does this and the previous part tell us about $S$? After that let $(x,y) : 1_T\to t$ be the unique morphism and note that $(xt,y)$ is a morphism from $t$ to $t$ and hence must be $(1_S,1_T)$ meaning that $xt=1_S$ and $y=1_T$. This means that $txt=t$ and hence $(1_S,tx)$ is a morphism from $t$ to $t$. Why does this mean that $tx=1_T$? What can you conclude?