# Adjunctions - Examples in which triangle identities fail

I’m on this journey of trying to understand what adjoint functors are, particularly interested in the formulation via unit and counit. If I’m not wrong, the triangle identities are encoding the naturality involved in the definition via the bijections, but I think I may be missing something.

In order to get a better understanding, I’m looking for examples where we have a pair of functors $$F \colon \mathcal{A} \to \mathcal{B}$$, $$G \colon \mathcal{B} \to \mathcal{A}$$ together with two natural transformations $$\alpha \colon \mathrm{id}_{\mathcal A} \Rightarrow GF$$, $$\beta \colon FG \Rightarrow \mathrm{id}_{\mathcal B}$$ in which one of the identities $$\beta F \cdot F \alpha = 1_F$$ or $$G\beta \cdot \alpha G = 1_G$$ doesn’t hold.

• You can take an equivalence of categories which is not an adjoint equivalence. Jan 25, 2023 at 9:18

You can take pretty much any "random" example of $$F$$, $$G$$, $$\alpha$$ and $$\beta$$ that you can write down and they probably will not satisfy the triangle identities; the only difficulty is coming up with natural transformations $$\alpha$$ and $$\beta$$ (since for many examples of $$F$$ and $$G$$ there may not exist any such natural transformations at all). For example, let $$F=G:\mathtt{Set}\to\mathtt{Set}$$ be the product functor $$X\mapsto X\times\{0,1\}$$. There is a natural transformation $$\alpha:id\Rightarrow FG$$ where $$\alpha_X:X\to X\times\{0,1\}^2$$ is given by $$\alpha(x)=(x,0,0)$$ and there is a natural transformation $$\beta:id\Rightarrow GF$$ where $$\beta_X:X\times\{0,1\}^2\to X$$ is the projection. Then, for instance, the composition $$(\beta F\cdot F\alpha)_X:FX\to FGFX\to FX$$ is the map $$(x,t)\mapsto (x,t,0,0)\mapsto (x,0)$$ which is not the identity.
For another example, you could take $$F=G=id:\mathtt{Ab}\to\mathtt{Ab}$$ and $$\alpha=\beta:id\Rightarrow id$$ is the natural transformation given by multiplication by $$2$$ on every abelian group. Then the compositions $$\beta F\cdot F\alpha$$ and $$G\beta\cdot\alpha G$$ will both be multiplication by $$4$$, which is not the identity.