Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am learning some category theory to help me with my area of research. I am trying to get familiar with the notion of adjunction. In some books I see the authors proving that two functors form an adjunction with just a comment, and this is something I find a bit difficult to follow. I will give an example. Suppose that $\mathcal{E}$ is a topos with small colimits and consider the functors $$\Gamma:\mathcal{E}\to \mathbf{Set},\\ \Gamma E=\text{Hom}_{\mathcal{E}}(1,E)$$ and $$\Delta:\mathbf{Set}\to \mathcal{E},\\ \Delta S=\coprod_{s\in S} 1$$ The author says "morphisms $\Delta S\to E$ in $\mathcal{E}$ clearly correspond to functions $S\to \Gamma E$ of sets, so that this functor $\Delta$ is left adjoint to $\Gamma$". I know that this correspondence is what needs to be proved to have an adjunction, but how is it so obvious that the correspondence holds? I have found many situations like this before...It is probably something to do with my mathematical maturity in this area, but any help on how to look correctly at this would be much appreciated.

share|cite|improve this question
The first point, it is often helpful try to prove these things by hand, to get a feeling for adjunction's. The other point is that two major families of adjunctions (and this does not encapsulate everything) are free-forgetful adjoint pairs and product-hom type pairs. This example looks like a free-forgetful pair. – Baby Dragon Feb 16 '14 at 9:23
A natural thing to do is this, but I dont know if it is correct... $$\text{Hom}_{\mathcal{E}}(\Delta S, E)=\text{Hom}_{\mathcal{E}}(\coprod_{s\in S} 1, E)\\ \cong \coprod_{s\in S}\text{Hom}_{\mathcal{E}}(1, E)\;\;\text{(is this true? can we take the coproduct outside?)}\\ \cong \text{Hom}_{\mathbf{Set}}(1,\coprod_{s\in S}\text{Hom}_{\mathcal{E}}(1, E)) \;\;\;\;\text{(because} \; \text{Hom}_{\mathbf{Set}}(1,\bullet)\cong \text{id}\;?)\\ \cong \text{Hom}_{\mathbf{Set}}(\coprod_{s\in S}1,\text{Hom}_{\mathcal{E}}(1, E))\;\;\text{(can we do this?)}\\ \cong \text{Hom}_{\mathbf{Set}}(S,\Gamma E)$$ – triwer23 Feb 16 '14 at 11:19
You cannot extract coproducts like that. They become products. – Zhen Lin Feb 16 '14 at 11:46
Where can I find out more about the properties of limits and colimits? Is there a modern treatment? Because most textbooks I have seen do not give actual examples... – triwer23 Feb 16 '14 at 14:44
Any textbook on category theory will discuss those properties. Wikipedia and the nLab are also a good start. – Ingo Blechschmidt Feb 17 '14 at 12:52

For this to be obvious, you must know that left adjoints preserve colimits. Given that, simply observe that every set $S$ can be expressed as a coproduct, viz $\coprod_{s \in S} 1$. Coproducts are colimits, so if $\Delta : \mathbf{Set} \to \mathcal{E}$ is a left adjoint, then we must have $\Delta S \cong \coprod_{s \in S} \Delta 1$; and if $\Delta$ is a left adjoint to $\Gamma$, then we must have $$\mathbf{Set} (1, \Gamma E) \cong \mathcal{E} (\Delta 1, E)$$ but $\mathbf{Set} (1, -) \cong \mathrm{id}$, so $$\mathcal{E} (1, E) \cong \mathcal{E} (\Delta 1, E)$$ and therefore $\Delta 1 \cong 1$.

My only advice is this: do not try to learn topos theory without first being comfortable with general category theory.

share|cite|improve this answer
Why is it enough to prove that $\mathbf{Set}(1,\Gamma E)\cong \mathcal{E}(\Delta 1,E)$ instead of $\mathbf{Set}(S,\Gamma E)\cong \mathcal{E}(\Delta S,E)$? And since we don't know that the functors are actually adjoint, how do you get the iso $\mathbf{Set}(1,\Gamma E)\cong \mathcal{E}(\Delta 1,E)$? ... I am a bit confused, it seems you are using properties of adjoints before we know that they are adjoints. – triwer23 Feb 16 '14 at 10:20
I'm working backwards – assume you have an adjoint, figure out what it must be, then prove that it is. – Zhen Lin Feb 16 '14 at 11:46
Thank you very much. – triwer23 Feb 16 '14 at 14:44

This isn't really special to topoi. You just have to remember the definition of a coproduct. It implies $$\hom_\mathcal{E}(\Delta S,E)=\prod_{s \in S} \hom(1,E).$$ And by the very definition of a product of sets, this identifies with $$\hom_{\mathsf{Set}}(S,\hom(1,E)).$$ Done. More generally, if $\mathcal{E}$ is any category with coproducts and $X \in \mathcal{E}$ is any object, then $\mathsf{Set} \to \mathcal{E}, ~S \mapsto \coprod_{s \in S} X$ (the copower) is left adjoint to $\hom_\mathcal{E}(X,-) : \mathcal{E} \to \mathsf{Set}$.

share|cite|improve this answer
Even though it seems very naturally true, I can't prove the second identification of the product with $\text{hom}_{\mathbf{Set}}(S,\text{hom}(1,E))$... Is it true that $\prod_{s\in S} X\cong \text{hom}_{\mathbf{Set}}(S,X)$ for any set $X$? – triwer23 Feb 17 '14 at 5:27
triwer23: Yes, that is true for any set $X$. Think about the description of $\prod_{s \in S} X$ as $S$-indexed families of elements of $X$. – Ingo Blechschmidt Feb 17 '14 at 12:51
I have written "by the very definition of a product of sets". So you should recap this definition and then everything will be clear. – Martin Brandenburg Feb 17 '14 at 18:25

Such a claim usually means that verifying the details is routine and straightforward (i.e., the steps to follow are clear, and there is no need for anything clever in order to establish the result). With time, you'll learn to agree with such statements and in some case quickly do the proof mentally. For now, if you really wish to understand what is going on, actually carry out the proof.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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