I am trying to understand the answer given to this question.

From what I understand they are saying that if there exists an $f$ such that $Ff = g$ then by naturality it must have the property that $GFf = Gg$ from which faithfulness of $G$ implies $Ff = g$.

The problem is that I don't see why we are guaranteed to have a morphism $f$ with this property. The definition of a natural transformation states that for every $f: X \rightarrow Y$ we have $\eta_Y \circ F(f) = G(f) \circ \eta_X$, so I do not see how we can use naturality unless we already know some $f$ exists with $Ff = g$.


1 Answer 1


You are misreading the answer. They aren't assuming such an $f$ exists: they are explaining that you can find the formula for such an $f$ by assuming it exists and figuring out what it must be. Namely if $f:X\to Y$ exists, it must be given by the formula $f=\eta_Y^{-1} \circ Gg \circ \eta_X$.

Now stop assuming that $f$ satisfies $Ff=g$ and simply define $f=\eta_Y^{-1} \circ Gg \circ \eta_X$. You can now prove directly from this formula (using naturality of $\eta$ several times) that $GFf=Gg$, and so $Ff=g$.

Here are the details of the proof that $GFf=Gg$. We have $$GFf=GF\eta_Y^{-1}\circ GFGg\circ GF\eta_X.$$ Now by naturality of $\eta$ (applied to the map $\eta_X:X\to GFX$), $GF\eta_X\circ\eta_X=\eta_{GFX}\circ\eta_X$. Since $\eta_X$ is an isomorphism, $GF\eta_X=\eta_{GFX}$. Similarly, $GF\eta_Y=\eta_{GFY}$, so we have $$GFf=\eta_{GFY}^{-1}\circ GFGg\circ \eta_{GFX}.$$

Finally, naturality of $\eta$ (applied to the map $Gg:GFX\to GFY$) gives $GFGg\circ\eta_{GFX}=\eta_{GFY}\circ Gg$, and so $$GFf=Gg.$$

  • $\begingroup$ Sorry, I don’t quite follow one of the steps you take here. I can see how you show that $GF\eta_Y=\eta_{GFY}$. However, I don’t quite see how it follows that $GF(\eta_Y^{-1})=\eta_{GFY}^{-1}$. Do functors preserve inverses, or am I missing some other obvious thing? Thanks. $\endgroup$ May 28, 2019 at 3:50
  • 1
    $\begingroup$ Yes, functors preserve inverses. $\endgroup$ May 28, 2019 at 3:52
  • $\begingroup$ Yes, of course. I’m not sure why I was under the impression that they don’t necessarily do so. Thank you! $\endgroup$ May 28, 2019 at 14:53

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