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I was reading here about adjoint functors, and I was following the construction of the right adjoint to a left adjoint functor, and I kept getting tripped up over showing that the resulting functor actually was a right adjoint, according to the universal morphism definition.

Namely, we define a functor $F:\mathcal{C}\leftarrow \mathcal{D}$ to be a left adjoint functor if for each object $X$ in $\mathcal{C}$, there exists a terminal morphism $\epsilon_X$ from $F$ to $X$. Similarly, a functor $G:\mathcal{C}\rightarrow \mathcal{D}$ is said to be a right adjoint functor if for each object $Y$ in $\mathcal{D}$ there exists an initial morphism $\eta_Y$ from $G$ to $Y$.

We can construct a functor $G:\mathcal{C}\rightarrow \mathcal{D}$ as follows: by hypothesis, for each $X$ in $\mathcal{C}$, there is $G_X$ such that $\epsilon_X:F(G_X)\rightarrow X$ is a terminal morphism. Next, letting $f:X\rightarrow Y$ be a morphism in $\mathcal{C}$, we have $\epsilon_X:F(G_X)\rightarrow X$ and $\epsilon_Y:F(G_Y)\rightarrow Y$, so that we have the morphisms $f\circ \epsilon_X: F(G_X)\rightarrow Y$ and $\epsilon_Y:F(G_Y)\rightarrow Y$; using the terminal property of the terminal morphism, there exists a unique morphism $G_f:G_X\rightarrow G_Y$ such that $\epsilon_Y \circ F(G_f)=f\circ \epsilon_X$; we then let $G(f)=G_f$. We finally define the functor $G:\mathcal{C}\rightarrow \mathcal{D}$ by $G(X)=G_X$ and $G(f)=G_f$.

However, in order to show that this $G$ is a right adjoint functor, I must show that for each $Y$ in $\mathcal{D}$ there is an initial morphism $\eta_Y$ from $Y$ to $G$.

My initial attempt was to construct $\eta_Y$ as follows: choose $X$ in $\mathcal{C}$ such that $G(F(G(X)))=Y$, and then take $\eta_Y=G(\epsilon_X):G(F(G(X)))\rightarrow G(X)$. However, I can't see why I can choose such an $X$.

Any help would be greatly appreciated.

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I have a related general question regarding universal morphisms (for each object) from adjoint functors $F$: In the third paragraph you say "by hypothesis, for each $X$ in...". Is it the case that an adjoint functor determines a unique universal morphism, or is it just the case that there is one? The example I have in mind is this: Say I know of the functor mapping to exponential objects (in ${\bf{Set}}$, say) and I characterize the functor mapping to the product by demanding it to be the adjoint to the previous one. Does this alone specify the universal morphism, i.e. the $\mathrm{eval}$ map? –  NikolajK Jun 26 at 7:22

2 Answers 2

up vote 2 down vote accepted

Everything is correct, except the last one. You are very close to the right solution.

You want to construct an initial morphism $\eta_Y\colon Y\to G(X)$. Take $X=F(Y)$. Then you have the terminal morphism $\epsilon_{F(Y)}\colon F(G_{F(Y)})\to F(Y)$ and the identity morphism $id_{F(Y)}\colon F(Y)\to F(Y)$. Using the terminal property, there exists a morphism $\eta_Y\colon Y\to G_{F(Y)}=G(F(Y))=G(X)$, such that $\epsilon_{F(Y)}\circ F(\eta_Y)=id_{F(Y)}$. The morphism $\eta_Y$ is initial.

Proof. Firstly, let's prove that $\epsilon\colon F\circ G\to I_{\mathcal{C}}$ is a natural transformation. Let $f\colon X\to X'$ be a morphism in $\mathcal{C}$. Then $f\circ\epsilon_X=\epsilon_{X'}\circ F(G(f))$ because of the terminal property of the morphism $\epsilon_{X'}$. Therefore, $\epsilon$ is natural. Secondly, let's prove that $\eta\colon I_{\mathcal{D}}\to G\circ F$ is a natural transformation. Let $g\colon Y\to Y'$ be a morphism in $\mathcal{D}$. We should prove that $G(F(g))\circ \eta_Y=\eta_{Y'}\circ g$. For that, by the terminal property of the morphism $\epsilon_{F(Y')}$, it suffices to prove that $$ \epsilon_{F(Y')}\circ F(G(F(g))\circ \eta_Y)=\epsilon_{F(Y')}\circ F(\eta_{Y'}\circ g). $$ But $$ \epsilon_{F(Y')}\circ F(G(F(g))\circ \eta_Y)=\epsilon_{F(Y')}\circ F(G(F(g)))\circ F(\eta_Y)=F(g)\circ\epsilon_{F(Y)}\circ F(\eta_Y)= $$ $$ =F(g)=\epsilon_{F(Y')}\circ F(\eta_{Y'})\circ F(g)=\epsilon_{F(Y')}\circ F(\eta_{Y'}\circ g), $$ by the naturality of $\epsilon$ and by the definition of $\eta$. Thus, $\eta$ is natural.

Let's prove that $\eta_Y\colon Y\to G(F(Y))$ is initial. Let $f\colon Y\to G(X')$ be a morphism in $\mathcal{D}$. To prove that $\eta_Y$ is initial, we should find a morphism $h\colon F(Y)\to X'$ in $\mathcal{C}$, such that $G(h)\circ \eta_Y=f$ and then prove that such morphism is unique. Take $h=\epsilon_{X'}\circ F(f)$. Firstly we should prove that $G(\epsilon_{X'}\circ F(f))\circ\eta_Y=f$. By naturality of $\eta$ we have: $$ G(\epsilon_{X'}\circ F(f))\circ\eta_Y=G(\epsilon_{X'})\circ G(F(f))\circ\eta_Y=G(\epsilon_{X'})\circ \eta_{G(X')}\circ f. $$ Therefore, by the terminal property of the morphism $\epsilon_{X'}$, it suffices to prove that $$ \epsilon_{X'}\circ F(G(\epsilon_{X'})\circ \eta_{G(X')}\circ f)=\epsilon_{X'}\circ F(f). $$ But $$ \epsilon_{X'}\circ F(G(\epsilon_{X'})\circ \eta_{G(X')}\circ f)=\epsilon_{X'}\circ F(G(\epsilon_{X'}))\circ F(\eta_{G(X')})\circ F(f)= $$ $$ =\epsilon_{X'}\circ\epsilon_{F(G(X'))}\circ F(\eta_{G(X')})\circ F(f)=\epsilon_{X'}\circ F(f), $$ by the naturality of $\epsilon$ and by the definition of $\eta$.

Finally, we should prove that such $h$ is unique. Let $G(h)\circ\eta_Y=f$. Then $$ h=h\circ\epsilon_{F(Y)}\circ F(\eta_Y)=\epsilon_{X'}\circ F(G(h))\circ F(\eta_Y)=\epsilon_{X'}\circ F(G(h)\circ\eta_Y)=\epsilon_{X'}\circ F(f). $$

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I seem to be blanking on how to show that $\eta_Y$ is initial. My initial thoughts went like this: if $f:Y\rightarrow G(X)$ is a morphism in $\mathcal{D}$, then if $\eta_Y$ is initial, then there exists $h:F(Y)\rightarrow X$ that satisfies $FG(h)=F(g)\circ \epsilon_{F(Y)}$. That doesn't do much, however. I'm sure I must reduce finding $h:F(Y)\rightarrow X$ to choose such an $h$, which will exist and be unique due to the fact that $\epsilon_X$ is a terminal morphism. Any further hints? –  Hayden Feb 26 at 23:55
    
@Hayden See proof. –  Oskar Feb 27 at 9:03
    
Wow, thank you! –  Hayden Feb 27 at 15:24

This is an alternative solution to Oskar's.

Having done a little more work with this, I know now that we can form the initial morphisms $\eta_Y$ from $Y$ to $G$ by essentially proving that the universality definition implies the Hom-set bijection definition. We define for each $X$ in $\mathcal{C}$ and $Y$ in $\mathcal{D}$ a map $\Phi_{Y,X}^{-1}:hom_{\mathcal{D}}(Y,G(X))\rightarrow hom_\mathcal{C}(F(Y),X)$ by the rule $\Phi_{Y,X}^{-1}(g)=\epsilon_X\circ F(g)$, where $g:Y\rightarrow G(X)$.

We now show that this is natural in both arguments: let $X,X'$ be objects in $\mathcal{C}$, $Y,Y'$ be objects in $\mathcal{D}$, and $f:X\rightarrow X'$, $g:Y'\rightarrow Y$. Then $$\Phi_{Y',X'}^{-1}(G(f)\circ h\circ g)=\epsilon_{X'}\circ F(G(f))\circ F(h)\circ F(g)=f\circ \epsilon_{X}\circ F(h)\circ F(g)=f\circ \Phi_{Y,X}^{-1}(h) \circ F(g).$$

Finally, we show that this is a bijection: it is injective because if $\epsilon_X\circ F(g)=\epsilon_X\circ F(g')$, then because $\epsilon_X$ is a terminal morphism, we know that $g$ is the unique morphism such that $\epsilon_X\circ F(g)=\epsilon_X\circ F(g')=\epsilon_X\circ F(g)$, showing that $g=g'$. It is surjective because if $f:F(Y)\rightarrow X$ is a morphism in $\mathcal{C}$, then because $\epsilon_X$ is a terminal morphism, we know that there exists $g: Y\rightarrow G(X)$ such that $\epsilon_X\circ F(g)=g$.

Thus, $\Phi^{-1}$ is a natural isomorphism $\Phi^{-1}: hom_\mathcal{D}(-,G-)\rightarrow hom_\mathcal{C}(F-,-)$.

Next, we take the family of bijections $\Phi:hom_\mathcal{C}(F-,-)\rightarrow hom_\mathcal{D}(-,G-)$ which is itself a natural isomorphism. Note that $\epsilon_X=\Phi^{-1}_{G(X),X}(id_{G(X)})$.

We may finally define $\eta_Y:= \Phi_{Y,F(Y)}(id_{F(Y)})$. Then we have $\Phi_{Y,X}(f)=G(f)\circ \eta_Y$. This is precisely what is necessary to say that $\eta_Y$ is an initial morphism.

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