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Let $\mathscr C$ be a category and $A,B\in Ob_\mathscr C$. I don't think it matters, but assume that $\mathscr C$ is locally small. I want to find a natural transformation between the covariant Hom-functors $Hom(A,–)$ and $Hom(B,–).$ I've tried a few things, but I'm still uneasy when I think about any of my attempts.

My best attempt (I think):

Let $X$ be any object of $\mathscr C$, and let $f:X \rightarrow Y$ be any morphism of $\mathscr C$. Fix a morphism $k:B\rightarrow A.$ Define $\tau_X:Hom(A,X)\rightarrow Hom(B,X)$ by $\tau_Xg=gk$ for all $g:A\rightarrow X.$ Define $\tau_Y:Hom(A,Y)\rightarrow Hom(B,Y)$ by $\tau_Yh=hk$ for all $g:A\rightarrow Y.$ Now we show that our $\tau_X,\tau_Y$ give us a commutative diagram: For all $g:A\rightarrow X$, $(Hom(B,f)\circ\tau_X)g=Hom(B,f)(gk)=f(gk)=$$(fg)k=\tau_Y(fg)=\tau_Y(Hom(A,f)g)=(\tau_Y\circ Hom(A,f))g$. Then $Hom(B,f)\circ\tau_X=\tau_Y\circ Hom(A,f).$ In other words, we have a commutive diagram. Therefore $\tau_X$ and $\tau_Y$ define a natural transformation $\tau$ between $Hom(A,–)$ and $Hom(B,–)$.

Here is why I'm still uneasy about the above attempt: We don't necessarily know that such a $k$ exists (right?); not every category has maps between each object.

JDivision
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    Right. And who said there had to be such a natural transformation? – Randall Feb 13 '19 at 03:18
  • I thought that my professor told me to construct such a natural transformation. I must have misunderstood his instructions. – JDivision Feb 13 '19 at 03:23
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    You are dangerously close to the Yoneda lemma. – Randall Feb 13 '19 at 03:25
  • Really? How so? I haven't studied the Yoneda lemma yet, so I will start reading on it tonight. – JDivision Feb 13 '19 at 03:32
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    It says that such natural transformations come from maps $k:B \to A$. (Of course, there may be no such maps, so there would no such natural transformations.) – Randall Feb 13 '19 at 03:32
  • Oh, wow. That's exciting! Thank you. – JDivision Feb 13 '19 at 03:36
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    Anyway, your attempt is right. The proof should really hinge on associativity of composition, which your argument does. – Randall Feb 13 '19 at 03:37
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    All right. What is the intuition behind not always having such a natural transformation if we have no map from B to A? If I draw out a finite category, I think I can see why there would be problems. This is what I'm thinking: Before we would apply such a $\tau$ we would have maps from A–>X being sent to maps A –>Y, and we would want $\tau$ to transform this into maps B–>X being sent to maps B–>Y, but if we don't have a map B–>A, we can't always do this? There's no way to 'slide' these morphisms. I hope that made sense. – JDivision Feb 13 '19 at 03:53
  • @JDivision If you have such a natural transformation, say it's $\tau$, then notice $\tau_A: \operatorname{Hom}(A,A) \to \operatorname{Hom}(B,A)$. There's definitely an element in $\operatorname{Hom}(A,A)$, the identity on $A$. Then $\tau_A(1_A) \in \operatorname{Hom}(B,A)$, so there is an arrow $B \to A$. The Yoneda Lemma builds on this idea, and says that arrows $B \to A$ are in (natural) bijection with natural transformations $\operatorname{Hom}(A,-) \to \operatorname{Hom}(B,-)$. – B. Mehta Feb 15 '19 at 13:04

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