- #1
dmuthuk
- 41
- 1
Given two left adjoints F,H:\mathcal{C}\to\mathcal{D} of a functor G:\mathcal{D}\to\mathcal{C}, how do we show that F and H are naturally isomorphic? This is my idea so far (I am working with the Hom-set defenition of adjunction):
We need to construct a natural isomorphism \alpha. So, for each x\in\mathcal{C}, I need a morphism \alpha_x:F(x)\to H(x). Suppose we are given the natural isomorphisms \varphi:\mbox{Hom}(F-,-)\to\mbox{Hom}(-,G-) and \psi:\mbox{Hom}(H-,-)\to\mbox{Hom}(-,G-). Then, I can simply let \alpha_x := \varphi_{x,Hx}^{-1}\circ\psi_{x,Hx}(1_{Hx}). But, I am stuck here. I don't know how to show that for a given morphism f:x\to y in \mathcal{C}, H(f)\circ\alpha_x = \alpha_y\circ F(f).
We need to construct a natural isomorphism \alpha. So, for each x\in\mathcal{C}, I need a morphism \alpha_x:F(x)\to H(x). Suppose we are given the natural isomorphisms \varphi:\mbox{Hom}(F-,-)\to\mbox{Hom}(-,G-) and \psi:\mbox{Hom}(H-,-)\to\mbox{Hom}(-,G-). Then, I can simply let \alpha_x := \varphi_{x,Hx}^{-1}\circ\psi_{x,Hx}(1_{Hx}). But, I am stuck here. I don't know how to show that for a given morphism f:x\to y in \mathcal{C}, H(f)\circ\alpha_x = \alpha_y\circ F(f).