# Prove categories equivalence using compositions of functors

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?

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

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.

• 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
– S. R
Commented Dec 27, 2019 at 5:52
• Wow, I was totally confused with the second row of the diagram. Now it's the correct one. sorry
– S. R
Commented Dec 27, 2019 at 7:09
• @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. Commented Dec 27, 2019 at 9:35