# Functor $L:\mathcal{A}\to\mathcal{B}$ is left adjoint to $R$ functor, then $L$ preserves all colimit.

This is a statement made in Weibel Chpt 2, Sec 6, Adjoint Functors and Left/Right Exactness.

2.6.10 Let functor $$L:\mathcal{A}\to\mathcal{B}$$ be left adjoint to a functor $$R:\mathcal{B}\to\mathcal{A}$$ with $$\mathcal{A,B}$$ arbitrary categories. Then $$L$$ preserves all \operatorname{colim}its.

My thought goes as the following. Suppose $$A=\operatorname{colim}(A_i)$$. Then one has $$\operatorname{Hom}_{\mathcal{B}}(L(\operatorname{colim}(A_i),B)\cong \operatorname{Hom}_{\mathcal{A}}(\operatorname{colim}(A_i),RB)\cong \operatorname{Hom}_{\mathcal{A^I}}(A_i,\Delta RB)=RHS$$ where $$\Delta$$ is the diagonal functor $$\Delta:i\in I\to RB$$.

Now $$\Delta R(B)=R\Delta(B)$$ for sure. Hence $$RHS\cong \operatorname{Hom}_{\mathcal{A^I}}(A_i,R\Delta B)\cong \operatorname{Hom}_{\mathcal{B}}(LA_i,\Delta B)\cong \operatorname{Hom}_{\mathcal{B}}(\operatorname{colim}(LA_i),B).$$ Hence $$\operatorname{Hom}_{\mathcal{B}}(L(\operatorname{colim}(A_i),-)\cong \operatorname{Hom}_{\mathcal{B}}(\operatorname{colim}(LA_i)),-)$$ naturally equivalent. Then I conclude $$L(\operatorname{colim}(A_i))\cong \operatorname{colim}(LA_i)$$.

$$\textbf{Q:}$$ Is above reasoning correct? I just used naively adjointness of functors and obvious commutativity of $$R\Delta=\Delta R$$.

• Looks ok to me. You used Yoneda at the end. But don't you need to show more than that $L(\text{colim}(A_i))\cong \text{colim}(LA_i)$ as $objects$? You should prove that if $(a,\lambda_i)$ is a colimit cone in $\mathcal A$ of the functor $A$ then $(La,L\lambda_i)$ is a colimit cone in $\mathcal B$ of the functor $LA$. – Matematleta Jun 10 at 22:04
• @Matematleta The indication of $La_i$ is colimit cone indicated by intermediate adjunction $Hom(LA_i,\Delta B)\cong Hom_B(colim(LA_i), B)$ in the middle. – user45765 Jun 10 at 22:14
• @Matematleta I think the colimit cone demonstration comes for free during the proof via adjunctions. – user45765 Jun 10 at 22:15
• It is not clear to me that this is the case. It's been some time since I did category theory, but I remember that Awodey does it the way you did, and I had a quibble with his proof too when I first saw it! – Matematleta Jun 10 at 22:19

Your reasoning is essentially correct. In fact, it's completely correct if your target category has all colimits. (Ignoring some minor typos, and things that aren't really expressed clearly in places.)

The only major issue with it is the final natural isomorphism.

The statement you want to prove is that

If $$J:I\to C$$ is a diagram in $$C$$ (i.e. a functor from our index category $$I$$ to $$C$$), and $$A=\newcommand\colim{\operatorname{colim}}\colim_C J$$ exists, and if $$L:C\to D$$ is left adjoint to $$R:D\to C$$, then the colimit of $$L\circ J$$ exists in $$D$$ and is $$L(A)$$.

You're currently assuming that $$\colim_D L\circ J$$ exists when you write $$\newcommand\Hom{\operatorname{Hom}}\Hom_{[I,D]}(L\circ J,\Delta_B)\simeq \Hom_D(\colim(L\circ J),B).$$ Then you're writing $$\Hom_D(L(\colim J),B)\simeq \Hom_D(\colim(L\circ J),B)$$ and concluding $$L(\colim J)\simeq \colim(L\circ J)$$ by the Yoneda lemma.

Instead, once you've proven $$\Hom_D(L(\colim J),B)\simeq \Hom_{[I,D]}(L\circ J,\Delta_B),$$ you can immediately conclude that $$L(\colim J)$$ is by definition a colimit of $$L\circ J$$, since it corepresents the functor $$X\mapsto \Hom_{[I,D]}(L\circ J, \Delta_X)$$. This way you're making it clear that we don't need to assume that the colimit of $$L\circ J$$ exists to prove that it must be $$L(\colim J)$$.

Side note

I also want to address Matematleta's comment on the question, concerning it being unclear where we get the morphisms forming the cocone from in this argument.

To me this part really makes me feel the beauty of the Yoneda lemma.

The morphisms determining the cocone are encoded in the natural transformation we obtain.

Let $$A=\colim J$$.

We have a natural isomorphism of functors in $$B$$, $$\Hom_D(L(A),B)\simeq \Hom_{[I,D]}(L\circ J,\Delta_B).$$ If we let $$B=L(A)$$, we have a natural isomorphism $$\Hom_D(L(A),L(A))\simeq \Hom_{[I,D]}(L\circ J, \Delta_{L(A)}).$$ The left hand set has a distinguished element, $$1_{L(A)}$$. This corresponds to a natural transformation $$\lambda : L\circ J \to \Delta_{L(A)}$$ on the right hand side.

In other words, for every object $$X\in J$$, we get a map $$\lambda_X : L(J(X))\to L(A)$$, and these maps commute with the maps in the diagram $$L\circ J$$, since $$\lambda$$ is a natural transformation. Thus our natural isomorphism encodes the cocone for us already.

The reason I say this makes me feel the beauty of the Yoneda lemma is that the construction I just gave is a special case of the statement in the Yoneda lemma that if $$F:D\to \mathbf{Set}$$ is a functor, then $$\operatorname{Nat}(h^X,F) \simeq F(X),$$ where $$h^A$$ is the contravariant Yoneda embedding of $$X$$.

• This is the clearest explanation I have read. +1. – Matematleta Jun 11 at 22:20

I guess it's no harder to do it from scratch: suppose $$\left \langle a,\lambda _{i} \right \rangle$$ is a colimit cocone on the functor $$A$$. Then $$\left \langle La,L\lambda _{i} \right \rangle$$ is a cocone on $$LA$$. Let $$\left \langle x,\mu _{i} \right \rangle$$ be any other cocone on $$LA$$. Then taking adjoints, we get a cocone on $$A$$, namely, $$\left \langle Rx,\mu _{i}^{*}\right \rangle$$. Since $$\left \langle a,\lambda _{i} \right \rangle$$ is a colimit cocone, we get an arrow $$\phi ^{*}:a\rightarrow Rx$$ unique with property that $$\phi ^{*}\circ \lambda _{i}=\mu ^{*}_{i}$$. Taking adjoints again, it is easy to see that $$\phi :La\rightarrow x$$ is the unique arrow making the required triangle commute, for if we write $$\begin{array}{ccc} \hom( a,Rx) & \leftarrow & \hom (La,x) \\ \downarrow & & \downarrow \\ \hom(A_i,Rx) & \leftarrow & \hom(LA_i,x ) \end{array}$$ and follow $$\phi$$ around the square, we see that $$\phi ^{*}\circ\lambda _{i} =(\phi\circ L\lambda _{i} )^{*}$$. But LHS of this is just $$\mu ^{*}_{i}$$, so that $$\phi\circ L\lambda _{i} =\mu_{i}$$ and $$\phi$$ is unique because $$\phi ^{*}$$ is.