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Denote the diagram:

$\require{AMScd}$ \begin{CD} C_{1} @>{G_{1}}>> C_2 @>{G_{2}}>> C_3\\ @VV{F_1}V @VV{F_2}V @VV{F_3}V\\ D_{1} @>{H_{1}}>> D_2 @>{H_{2}}>> D_3\\ \end{CD}

Where $G_2\circ G_1, H_2\circ H_1,F_2$ are equivalences of categories. I need to prove that $F_1,F_3$ are also equivalences. It's easy to show that $G_1,H_1$ are faithfull, and therfore $F_1$ also, and similarly, $G_2,H_2$ are essentially surjective, and therefore $F_3$. Diagram chasing gives the opposite - $F_1$ is essentially surjective and $F_3$ is faithfull. I couldn't prove that $F_1,F_3$ are full. Any ideas?

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  • $\begingroup$ It is kind of implicit from the rest of your question, but you should say that the diagram commutes. $\endgroup$ Commented Dec 26, 2019 at 23:21

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First we prove that $F_1$ is full. Let $f: F_1(A) \to F_1(B)$, then $H_1(f): F_2 G_1(A) \to F_2 G_1(B)$. So since $F_2$ is full, we find $f': G_1(A) \to G_1(B)$ with $F_2(f') = H_1(f)$. Now we consider $G_2(f'): G_2 G_1(A) \to G_2 G_1(B)$, and using fullness of $G_2 G_1$ we find $f'': A \to B$ such that $G_2 G_1(f'') = G_2(f')$. Then we chase this construction back: $$ H_2 H_1 F_1(f'') = F_3 G_2 G_1(f'') = F_3 G_2(f') = H_2 F_2(f') = H_2 H_1(f), $$ so $F_1(f'') = f$ by faithfulness of $H_2 H_1$, and we conclude that $F_1$ is full.

To complete the entire argument we can indeed use that $F_3$ is isomorphic to a composition of equivalences (and thus itself an equivalence). We can also use a (not too hard) elementary proof to see that $F_3$ is full.

To see that, let $g: F_3(X) \to F_3(Y)$. Then because $G_2 G_1$ is essentially surjective, there are $X', Y'$ such that $G_2 G_1(X') \cong X$ and $G_2 G_1(Y') \cong Y$. So we have an arrow $$ \bar{g}: F_3 G_2 G_1(X') \xrightarrow{F_3(\cong)} F_3(X) \xrightarrow{g} F_3(Y) \xrightarrow{F_3(\cong)} F_3 G_2 G_1(Y'). $$ Since the isomorphisms are in the image of $F_3$, it suffices to show that $\bar{g}$ is in the image of $F_3$. The domain and codomain of $\bar{g}$ are the same as $H_2 H_1 F_1(X')$ and $H_2 H_1 F_1(Y')$ respectively. So using fullness of $H_2 H_1$ and $F_1$ (as just established), we find $g': X' \to Y'$ such that $$ F_3 G_2 G_1(f') = H_2 H_1 F_1(f') = \bar{g}, $$ and we conclude that $F_3$ is full.

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  • $\begingroup$ Why is it easy to show that $H_2$ is full? And about $F_1$ it's easier: $F_1 = (H_2 \circ H_1)^{-1} \circ F_3 \circ (G_2 \circ G_1)$, which is a composition of category equivalences $\endgroup$
    – S. R
    Commented Dec 27, 2019 at 5:52
  • $\begingroup$ Wow, I was totally confused with the second row of the diagram. Now it's the correct one. sorry $\endgroup$
    – S. R
    Commented Dec 27, 2019 at 7:09
  • $\begingroup$ @S.R You are right, I edited with a better (correct) argument. As per your second question: $F_1$ is not quite equal to that composition, $H_2 H_1$ only has an inverse up to isomorphism (we are talking about equivalences, not isomorphisms of categories). But that means that $F_1$ is isomorphic to the mentioned composition. I added a bit about that in my answer. $\endgroup$ Commented Dec 27, 2019 at 9:35

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