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Hi, I apologise if this is the wrong place for this question but i need to ask it somewhere. The question is whether the right, (perhaps left?), adjoint to $Hom(-,A)$ exists in $\bf Set$ and how to deduse it.

I would very much like it to be the disjoint union $-\oplus A $ but im not quite sure how to deduce it or if its simply whishfull thinking.

Alternativly, is there an other adjoint (left or right) to the disjoint union.

I am writing an undergrad paper and I am aware of the coproduct. I'm asking since i hope the deduction of a dual adjoint will tell me a little of the general workings of category theory and help me understand how to make "computations" in concrete categories.

Also, if you have the time and will, is there a right adjoint to the covariant Hom functor?

Thanks in advance or sorry for wasting your time

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3 Answers 3

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To speak of right (or left) adjoints for a contravariant functor $F: C\to D$, one needs to decide whether to view it as a functor from $C^{op}$ to $D$ or as a functor from $C$ to $D^{op}$. What the one viewpoint calls a left adjoint, the other will call a right adjoint. One therefore often speaks instead of two contravariant functors being "adjoint on the right" (or on the left). In this language, $Hom(-,A)$ is adjoint on the right to itself, which means that morphisms from $X$ to $Hom(Y,A)$ are in natural bijective correspondence with morphisms from $Y$ to $Hom(X,A)$; here "natural" means (as usual for adjointness) with respect to both $X$ and $Y$.

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    $\begingroup$ Thx, I appreaciate all answers are good. what really had me confused was the contravariant nature and my own hopes. $\endgroup$
    – Tobias
    Commented Feb 19, 2012 at 23:08
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Left adjoints preserve colimits, but $\mathrm{Hom}(-,A)$ does not because $\mathrm{Hom}(\varnothing,A)\neq\varnothing$.

Right adjoints preserve limits, but if $A$ is not terminal, $\mathrm{Hom}(-,A)$ does not because $\mathrm{Hom}(\{*\},A)\neq\{*\}$. (if $A$ is terminal, the functor is constant equal to $\{*\}$ and has a left adjoint which is the constant functor equal to $\varnothing$).

You can do the same reasoning to deduce that $-\oplus{}A$ does not have adjoints, unless $A=\varnothing$.

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  • $\begingroup$ Nice explanation. $\endgroup$
    – Samuel Vidal
    Commented Feb 19, 2012 at 20:22
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    $\begingroup$ You forgot the contravariance. Whichever hand it is, we do have $\hom(\emptyset,A) = \lbrace\ast\rbrace$, so we're preserving that limit, at least. $\endgroup$
    – Theo Johnson-Freyd
    Commented Feb 20, 2012 at 2:44
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Covariant $Hom$ takes $x$-element sets to $x^n$-element sets. If $f$ is a right adjoint, we have $y^{x^n}=f(y)^x$, and there is no function with this property.

But product is left adjoint to covariant $Hom$, as follows:

$Hom(X,Hom(A,Y))=Hom(X \times A,Y)$

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