# Polynomial Functor is ω-Continuous

Suppose $P:Set\rightarrow Set$ is given by $X \mapsto \sum_{i=0}^{n} C_{i}\times X^{i}$. I want to prove that $P$ is ω-continuous. Three questions: 1). Is my following argument correct? 2). Is there an explicit construction using the product and disjoint union operations? 3). What can be said if $Set$ is replaced by an arbitrary category $C$, say one that is not Cartesian Closed?

edit: I cheated and found a proof of a more general result in CWM that takes care of this. But won't the following argument work for $Set$?

1). Constant endofunctors of Set are ω-continuous.

2). The diagonal functor $\Delta :Set\rightarrow Set\times Set\cdots \times Set$ is left adjoint to the corresponding product functor and so preserves colimits. Since the objects in $Set\times Set\cdots \times Set$ are the categorical products in $Set$, this says if $X_{i}\rightarrow X$ then $X_{i}^{n}\rightarrow X^{n}$.

3). The composition of ω-continuous functors is ω-continuous.

4). The coproduct $+ : Set\times Set → Set$ is left adjoint to $X\mapsto (X, X)$.

Thanks in advance for taking the time to review this very basic stuff.

The following was my attempt to show item 2). from scratch.

I show that if $X_{i} \mapsto X$, then $(X_{i}\times X_{i}) \mapsto (X \times X)$. I am having trouble with the last step. Any help would be greatly appreciated. Here's what I have so far:

Let $\left \langle X_{i},\alpha _{i} \right \rangle$ be a colimit cone for $\Delta :\omega \rightarrow Set$. We then have maps $f_{ij}:X_{i}\rightarrow X{j}$ s.t $\alpha _{j}\cdot f_{ij}=\alpha _{i}$

Consider the diagram $:\omega ^{2}\rightarrow Set$ given by $(i,j) \mapsto X_{i}\times X_{j}$. Then we get the commuting squares:

\begin{matrix}X_{i}\times X_{j} &\xrightarrow{(id,f_{j,j+1})}& X_{i}\times X_{j+1} \\ \ \ \downarrow (f_{i,i+1},id) & & \ \ \downarrow (f_{i,i+1},id) \\ X_{i+1}\times X_{j} &\xrightarrow{(id,f_{j,j+1})}& X_{i+1}\times X_{j+1} \end{matrix}

Next, I proved that

$X_{m}\times X_{k} \mapsto X\times X_{k}$ for each $k$. This gives us a new diagram $\omega \rightarrow Set$ which is of course $X\times X_{k}$. This diagram has colimit $X\times X$.

In proving the above, I just observed that, for fixed $k$, if $\beta_{m}: X_{m} \times X_{k}\rightarrow Z$ are arrows making the diagram commute, the we get arrows $\overline{\beta_{m}}:X_{m}\rightarrow Z$ making the original diagram commute, which then give us a unique morphism $\overline{\phi}:X\rightarrow Z$, satisfying $\overline{\phi}\cdot \alpha _{k}=\overline{\beta_{k}}$. This then gives us the unique $\phi:X\times X_{k}\rightarrow Z$. (Or we could just say the functors we get for each $k$, namely, $-\times X_{k}$, peserve colimits, being left adjoints).

A similar argument works for the second colimit.

Now, I want to say that $X\times X$ is the limit of the diagonal diagram $X_{i}\times X_{i}$. So, suppose $\beta_{i,i}:X_{i}\times X_{i}\rightarrow Z$, is a cocone for the diagonal diagram. The fact that all the squares commute induces a cocone $\beta_{j,k}:X_{j}\times X_{k}\rightarrow Z$ over the whole diagram, and in particular if we fix $k$, we obtain unique $\phi_{k}:X\times X_{k}\rightarrow Z$ s.t. $\phi_{k}\cdot \left \langle \alpha _{j},id_{k} \right \rangle=\beta_{j,k}$

I want to use this to give me the required arrow $:X_{i}\times X_{i}\rightarrow Z$, but I do not see how.

• Who says $X \mapsto X \times X$ preserves colimits? It doesn't. The claim is not true in absolute generality – it is true for $\mathbf{Set}$ and more generally in $\sigma$-pretoposes, but there is work to be done! – Zhen Lin Jun 28 '14 at 18:31
• @Zhen: In $Set$, I fix $X=\varinjlim X_i$, then $\varinjlim (X_i\times X)=X\times X$ because $-\times X$ is a left adjoint. What am I missing? – Matematleta Jun 28 '14 at 18:46
• Sure, $X \mapsto X \times Y$ preserves colimits. That means nothing for $X \mapsto X \times X$. – Zhen Lin Jun 28 '14 at 19:03
• @Zhen: Yes, I see what you mean now. I need to show $X_{i} \mapsto X \Rightarrow (X_{i}\times X_{i}) \mapsto (X\times X)$.How about this: I consider $Set \times Set$ as functors from $\textbf{2}$ to $Set\times Set$ so the object $(x,x)$ is the functor $0 \mapsto x$ and $1 \mapsto x$. Then I compute the colimit pointwise. Or is it just as easy to do this from scratch? – Matematleta Jun 28 '14 at 19:37
• I don't know what you're trying to do. There's no point trying to prove that $X \mapsto X \times X$ preserves colimits – it doesn't: $1 \times 1 + 1 \times 1 \ne (1 + 1) \times (1 + 1)$. But it does preserve filtered colimits. – Zhen Lin Jun 28 '14 at 21:47