Natural isomorphism of Left adjoints

  • Thread starter dmuthuk
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  • #1
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Main Question or Discussion Point

Given two left adjoints [tex]F,H:\mathcal{C}\to\mathcal{D}[/tex] of a functor [tex]G:\mathcal{D}\to\mathcal{C}[/tex], how do we show that [tex]F[/tex] and [tex]H[/tex] 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 [tex]\alpha[/tex]. So, for each [tex]x\in\mathcal{C}[/tex], I need a morphism [tex]\alpha_x:F(x)\to H(x)[/tex]. Suppose we are given the natural isomorphisms [tex]\varphi:\mbox{Hom}(F-,-)\to\mbox{Hom}(-,G-)[/tex] and [tex]\psi:\mbox{Hom}(H-,-)\to\mbox{Hom}(-,G-)[/tex]. Then, I can simply let [tex]\alpha_x := \varphi_{x,Hx}^{-1}\circ\psi_{x,Hx}(1_{Hx})[/tex]. But, I am stuck here. I don't know how to show that for a given morphism [tex]f:x\to y[/tex] in [tex]\mathcal{C}[/tex], [tex]H(f)\circ\alpha_x = \alpha_y\circ F(f)[/tex].
 

Answers and Replies

  • #2
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You have (natural) isos

(F?,?)-->(?,G?)-->(H?,?)

is the composition of (natural) isos an iso?
 
  • #3
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You have (natural) isos

(F?,?)-->(?,G?)-->(H?,?)

is the composition of (natural) isos an iso?
Yes, I believe so.
 

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