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Let us suppose that $C$ is a category where small limits are representable and $D$ is a small category.

Let $F:Hom(D,C)\rightarrow C$ be the functor defined by $F(f)=\varprojlim f$.

Is it true that $F$ has a right adjoint constant functor? Can someone please explain this to me? What about the unit and co-units here?

Also, could someone please suggest me some nice introductory books on this stuff? I am finding it very confusing.

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Thanking you in advance. –  nicole Dec 10 '13 at 19:33

2 Answers 2

No, $F$ has no right adjoint in general, but it has a left adjoint, which sends each object $c$ of $C$ to the constant functor in $\text{Hom}(D,C)$ that sends every object of $D$ to $c$.

I'm probably old-fashioned, but I still think the best introductory book on category theory is Saunders Mac Lane's "Categories for the Working Mathematician".

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I like Mac Lane as well, though I also like Awodey's Category Theory, which doesn't cover as much ground, but is much friendlier to newcomers. –  Malice Vidrine Dec 10 '13 at 20:19

As Andreas already answered, $F$ has no right adjoint in general. Let's see why this is so on a simple example.

Take $C=\mathrm{Set}$, the category of sets and take $D$ a category with two objects, call them $*$ and $**$, and only identity morphisms. Then given an element $f \in \mathrm{Hom}(D, C)$ its limit is simply the product $f(*) \times f(**)$. Now limits and colimits in $\mathrm{Hom}(D, C)$ are given pointwise. Take $f, g : D \to C$ to be the constant $1$ (terminal object in $\mathrm{Set}$) functors. Then $F(f + g) = (1 + 1) \times (1+1)$ which is a set of cardinality $4$. On the other hand $F(f) + F(g) = 1 \times 1 + 1 \times 1$ which is a set of cardinality $2$, thus $F(f + g) \not\cong F(f) + F(g)$.

Now if $F$ were to have a right adjoint it would itself be a left adjoint and left adjoints preserve colimits. $F$ does not preserve colimits, thus it has no right adjoint.

Note however that in a quite a few cases $F$ will preserve colimits and will have a right adjoint.

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Thank you @ Andreas and @Ales, this makes good sense. But just now I have found another question in this website ( titled Problem on Adjoint Functor by Mr Andy Flower) which is very similar to mine, except that Mr Flower wants to prove that $F$ has a right adjoint k such that k(X) is the constant functor under the same setting. What is your take on this one? Does it not imply that right adjoint exists? Or is that question false? –  nicole Dec 11 '13 at 6:39
    
Well the question is wrong in the sense that what they want does not exist. In general, there will be no right adjoint. If this is homework then whomever stated it made a mistake. –  Aleš Bizjak Dec 11 '13 at 7:27
    
Note however that if you were to take $C$ to have small colimits and $F$ to be the colimit functor, then it would have the constant functor functor as the right adjoint. –  Aleš Bizjak Dec 11 '13 at 7:29
    
Thank you so much. –  nicole Dec 11 '13 at 15:35

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