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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 $f' \circ a = b \circ f$

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    $\begingroup$ Consider what the final objects in $\mathbf{Set}$ are; this should give you a clue as to what the final objects in this category should be. $\endgroup$ – Clive Newstead Aug 19 '19 at 20:27
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    $\begingroup$ If we write $\to$ for the category consisting of two objects and one arrow joining them (and the identities), then the arrow category of $\mathcal C$ is $\mathcal C^{\to}$, i.e. the category of functors from $\to$ to $\mathcal C$. Now you can apply generic theorems about limits in functor categories, in particular that they are computed point-wise when the target category is complete. You are probably more at a point where you should just try to directly prove the statement though... $\endgroup$ – Derek Elkins left SE Aug 19 '19 at 21:00
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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?

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