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• CommentRowNumber1.
• CommentAuthorMirco Richter
• CommentTimeMar 3rd 2012
• (edited Mar 3rd 2012)

Is there a well defined internal hom for cosimplicial objects? (sets, algebras, rings)

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeMar 3rd 2012
• (edited Mar 3rd 2012)

See my response here, which works in any finitely complete (edit: and cocomplete) cartesian closed category $C$ (which includes cosimplicial sets, simplicial sets, and much more). The equalizer of the diagram gives you the hom of internal $T$-algebras as an object of $C$.

If you want the internal hom of two internal $T$-algebras as another internal $T$-algebra, you don’t get that unless the algebraic theory in question is a commutative algebraic theory. In that case, you use the same equalizer, and it acquires a $T$-algebra structure by using commutativity of $T$.

• CommentRowNumber3.
• CommentAuthorMirco Richter
• CommentTimeMar 3rd 2012
• (edited Mar 3rd 2012)

Yes I was aware of your response, but likely someone has already worked it out explicitly in the cosimplicial situation, so I don’t have to do it by myself.

Having a general way to define the internal hom is surely of great meaning, but I don’t want to reinvent the wheel in this particular situation.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeMar 3rd 2012

Mirco, why did you ask is there an internal hom, if as you say you were aware of what I (and others) had written? I was simply answering the question as stated.

By the way, this description is explicit, if you know how to form exponentials in a presheaf topos. If $X, Y: C^{op} \to Set$ are presheaves on $C$, then the value of $Y^X: C^{op} \to Set$ at an object $c$ is the set

$Y^X(c) = Nat(X \times C(-, c), Y)$

It’s probably a good exercise to work out the remaining details of the construction; I’m not sure where you can find it written down in a text. Possibly the details should be “Labbified” by somebody, but I do encourage you to give it a go on your own first, because these are good techniques to be familiar with.

• CommentRowNumber5.
• CommentAuthorMirco Richter
• CommentTimeMar 4th 2012
• (edited Mar 4th 2012)

Ok the ’pure logic’ correct answer to my initial question so is ’yes’ as you stated and that’s all to say. I agree formally. But in my opinion answering the question with the explicit construction (as you did in #4.Thanks!) is the better choice, simply because this is an open forum and the approach that has the “minimum overhead” is likely to be the most useful to the readers (that are often not experts in these kind of things).

Nevertheless I wonder about the explicit equation for $Y^X(c)$ in case of cosimplicial sets because they are covariant functors on $\Delta$ and hence their representables are $Hom([n],.)$ ?

Does $Hom(X, Y \times Hom([n],.))$ makes any sense instead?

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeMar 4th 2012

Just take $C = \Delta^{op}$, where presheaves are cosimplicial sets. I was just saying how it works for a general presheaf topos. Alternatively, to give the construction for copresheaves $X, Y: C \to Set$, you’d switch to $Y^X(c) = Nat(X \times \hom(c, -), Y)$. Same thing, either way.

• CommentRowNumber7.
• CommentAuthorMirco Richter
• CommentTimeMar 4th 2012

ok

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeMar 4th 2012
• (edited Mar 4th 2012)

You might find the following additional hints useful. Let us take the theory of rings as our example. So, if $X$ and $Y$ are internal ring objects, there are operations $+: X^2 \to X$, $\cdot: X^2 \to X$, $0: 1 \to X$, $e: 1 \to X$ where $1$ is the terminal presheaf and $e$ denotes the multiplicative unit, and similarly for $Y$.

For each object $c$ of $C$, we want the equalizer subobject of $Y^X$ that expresses the defining equations for a ring homomorphism. We can do this succinctly and elegantly by the construction I mentioned in the other thread (which involves the free object $T X$ that bundles all operations together in one blow), but for calculational purposes it might be more concrete to focus one on operation at a time. So suppose we take just the equation that says a homomorphism must preserve multiplication.

We are trying to internalize the equational condition on a transformation $f: X \to Y$ that the two paths around the square

$\array{ X^2 & \stackrel{\cdot_X}{\to} & X \\ ^\mathllap{f^2} \downarrow & & \downarrow^\mathrlap{f} \\ Y^2 & \underset{\cdot_Y}{\to} & Y }$

give equal maps. Internally, we are thus equalizing two maps of the form $Y^X \stackrel{\to}{\to} Y^{X^2}$. The first,

$Y^{(\cdot_X)}: Y^X \to Y^{(X^2)},$

pulls back $f$ along multiplication $\cdot_X$. The second is a composite

$Y^X \to (Y^2)^{(X^2)} \stackrel{(\cdot_Y)^{(X^2)}}{\to} Y^{(X^2)}$

where the first map internalizes the functor that takes $f$ to $f^2$, and it is this map we need to describe more explicitly in terms of the formula for exponentials I gave you before.

So we now need to describe explicitly the $C$-natural family of functions

$Nat(X \times \hom(-, c), Y) \to Nat(X^2 \times \hom(-, c), Y^2).$

It shouldn’t be too hard to guess. It’s the composite

$Nat(X \times \hom(-, c), Y) \to Nat(X^2 \times \hom(-, c)^2, Y^2) \to Nat(X^2 \times \hom(-, c), Y^2)$

where the first arrow essentially takes a transformation to its cartesian square, and the second involves pulling back along a diagonal map $\hom(-, c) \to \hom(-, c) \times \hom(-, c)$. It’s the second which involves the concept of canonical strength on a functor (in this case the cartesian square functor) which is discussed over at commutative algebraic theory and, further back, at tensorial strength. It’s sort of the key to the construction.

Putting all this together, the equalizer condition amounts to taking, for each object $c$, the equalizer of two functions of the form

$Nat(X \times \hom(-, c), Y) \stackrel{\to}{\to} Nat(X^2 \times \hom(-, c), Y)$

(whose definitions should be clear at this point). That’s just for multiplication, of course; there are three other operations to consider. The intersection of the four equalizer subsets gives you the $C$-natural family which is the desired internal object $\hom_{Ring}(X, Y) \hookrightarrow Y^X$.

I hope at this point you have everything you need, so that you are now free to explore the nitty-gritties to your heart’s content. My earlier answers were much more succinct because it was not clear how much detail you required. (Plus, it takes time to type up more detailed answers.) That by the way is a plea to make your questions more explicit in the future, so that one doesn’t have to guess at what you need. :-)

• CommentRowNumber9.
• CommentAuthorMirco Richter
• CommentTimeMar 5th 2012

I need some time to get through it…