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.

  • $\begingroup$ 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? $\endgroup$ – Nikolaj-K Jun 26 '14 at 7:22

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 $$ h=h\circ\epsilon_{F(Y)}\circ F(\eta_Y)=\epsilon_{X'}\circ F(G(h))\circ F(\eta_Y)=\epsilon_{X'}\circ F(G(h)\circ\eta_Y)=\epsilon_{X'}\circ F(f). $$

  • $\begingroup$ 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? $\endgroup$ – Hayden Feb 26 '14 at 23:55
  • $\begingroup$ @Hayden See proof. $\endgroup$ – Oskar Feb 27 '14 at 9:03

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.