Emily Riehl's "Category Theory in Context", ${\rm Exercise}~2.1.{\rm ii}.$

Prove that if $F:{\rm C}\to{\rm Set}$ is representable, then $F$ preserves monomorphisms, i.e., sends every monomorphim in ${\rm C}$ to an injective function. Use the contrapositive to find a covariant set-valued functor defind on your favorite concrete category that is not representable.

I have a rough idea how to approach, but $a)$ I am not sure if my reasoning is valid and $b)$ how to justify one step. Anyway, lets start.

Keep in mind that $f$ is a monomorphism if it is left cancellable, i.e $f\circ x=f\circ y\implies x=y$. Given a representation of $F:{\rm C}\to{\rm Set}$ by an object $c$ and thus a natural isomorphism $F\cong{\rm Hom}(c,-)$ it follows that \begin{align*} f\circ x&=f\circ y\tag1\\ F(f\circ x)&=F(f\circ y)\tag2\\ \cong~{\rm Hom}(c,f\circ x)&={\rm Hom}(c,f\circ y)\tag3\\ {\rm Hom}(c,x)&={\rm Hom}(c,y)\tag{$\color{red}4$}\\ \cong F(x)&=F(y)\tag5 \end{align*} Line $(3)$ and line $(5)$ follow from the given natural isomorphism. What bothers me is line $(4)$ (as emphasized above). I am not entirely sure if one can argumentate like this. As $f$ is a monomorphism this property is inherited to hom-sets involving $f$. Anyway, I do not know how to proper phrase this step (if it's even possible to do so!).

Is my proof correct, if so how to justify line $(4)$; if not, please explain what went wrong. Additionally, I would like to see an example for the second part of the exercise as I cannot really think of one in particular.

Thanks in advance!

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    $\begingroup$ What is $C(c, f \circ x)$? $F$ being representable means that $F \simeq Hom(X, -)$, so I take it that your $C(-, -)$ notation is means the standard $Hom$-set. However, both arguments then need to be objects of your category, but $f \circ x$ is a morphism. $\endgroup$ – xyzzyz Sep 30 '19 at 18:18
  • $\begingroup$ @xyzzyz Yes, $C(-,-)$ is the usual Hom-set. I'll edit to avoid ambiguity. I guess I messed up the definition of a representation then... Could you elaborate? $\endgroup$ – mrtaurho Sep 30 '19 at 18:21
  • $\begingroup$ I don't think your edits helped much. What morphisms are there between an object $c$ and a morphism $f \circ x$? $\operatorname{Hom}(c, f \circ x)$ is still an undefined object. $\endgroup$ – xyzzyz Sep 30 '19 at 18:28
  • $\begingroup$ @xyzzyz I see the issue, and I guess my edits clarify that I was thinking in a totally wrong manner. Wouldn't it be enought to specify that $x$ and $y$ are objects? $\endgroup$ – mrtaurho Sep 30 '19 at 18:29
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    $\begingroup$ If $x$ is an object, what is $f \circ x$ then? $\endgroup$ – xyzzyz Sep 30 '19 at 18:30

I think you went too deep following a wrong track of reasoning, and now are very much confused as to what objects you are even dealing with. There isn't much to the problem if you approach it correctly.

Suppose $f: X \to Y$ is mono. We want to show that $F(f): F(X) \to F(Y)$ is injective function of sets. Since $F$ is representable, $F$ is naturally isomorphic to $\operatorname{Hom}(A, -)$ for some $A \in \operatorname{Ob}(\mathbf{C})$. This means that we have the following commutative diagram in the category of sets:

$\require{AMScd}$ \begin{CD} F(X) @>{F(f)}>> F(Y)\\ @V{\mu_X}VV @V{\mu_Y}VV\\ \operatorname{Hom}(A, X) @>{f \circ -}>> \operatorname{Hom}(A, Y) \end{CD}

where $\mu_X, \mu_Y$ are isomorphisms (bijections), and we need to prove that $F(f)$ is injective. Since the vertical arrows are bijections, to prove that the top arrow is injective, it's enough to prove that the bottom arrow is injective. What it means for bottom arrow to be injective? Take any $x, y \in \operatorname{Hom}(A, X)$. For the bottom arrow $f \circ -$ to be injective, we need to have that whenever $f \circ x = f \circ y$, necessarily $x = y$. But that follows immediately from $f$ being monomorphism.

  • $\begingroup$ Ah, yes. Thank you kindly. Indeed, I totally confused myself while switching objects and morphisms around. I've had the idea (I guess) but I wasn't able to properly apply it. $\endgroup$ – mrtaurho Sep 30 '19 at 18:46
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    $\begingroup$ Why is it that the bottom arrow implies the top arrow is injective? Is it because if $f\circ-\mu_X$ is inj, then $\mu_YFf$ has the same properties so is inj, which implies $Ff$ is inj? So if I had drawn the diagram with the isos facing the other direction, it would not have worked? $\endgroup$ – user524154 Sep 30 '19 at 19:13
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    $\begingroup$ If top arrow glues two points together as you go right, these are also glued as you go right and then down, but by commutativity this is the same as going first down and then right, but going down is a bijection, so if any gluing happens, it must be on the bottom arrow, which can't happen if it's injective. $\endgroup$ – xyzzyz Sep 30 '19 at 20:49

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