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I am trying to prove the following: Let $C$ be a category with an initial object, and ω-limits. Suppose $F$ is an ω-continuous endofunctor on $C$. Then $F$ has an initial algebra. I know the result is well-known but I want to do it on my own, with a few hints. Here is what I have so far:

I started with an $\Delta : ω\rightarrow C$, of the form $0\rightarrow F0\rightarrow F^{2}0\rightarrow \cdot \cdot \cdot \cdot \rightarrow F^{n}0\rightarrow \cdot \cdot \cdot $, with limit $A$, and found an isomorphism $i:FA\rightarrow A$ by showing that $FA$ is another colimit for the diagram $\Delta k$, of which $A$ is also still a colimit, where $k$ is just the map which sends $n$ to $n+1$. Then I let $x:FX\rightarrow X$ be any algebra. I was able to construct a cocone to $X$ under $\Delta $ by using the fact that $C$ has an initial object, from which I join together the appropriate commuting squares. This, in turn gives me an $f:A\rightarrow X$ satisfying the required UMP. I then showed that

\begin{array}{ccc} FA & \xrightarrow{Ff} & FX \\ \downarrow i & & \downarrow x \\ A & \xrightarrow{f} & X \end{array}

commutes.

edit: Now in order to prove that $i:FA\rightarrow A$ is initial I show that this $f$ is the only arrow that makes the square commute, using Zhen's hint in the comment section: induction. So, let $(A,\delta _{n})$ be the colimit cocone for $\Delta $ and $(X,\xi _{n})$ the abovementioned cocone to $X$. Then

$f\cdot \delta _{n}=\xi _{n}$ and $x\cdot Ff=f\cdot i$ and $i$ is an isomorphism. Let $g:A\rightarrow X$ be any morphism of $F$-Algebras. We show that $g\cdot \delta _{n}=\xi _{n}$ for all $n$:

This just amounts to gathering the following facts:

1). $g\cdot \delta _{0}=\xi _{0}$ (trivial).

2). $(FA,F\delta _{n})$ is a colimit cocone ($F$ preserves colimits).

3). $i\cdot F\delta _{n}=\delta _{n+1}$ (by 2. Actually, $i$ was constructed this way).

4). $x\cdot F\xi _{n}=\xi _{n+1}$ (by construction of $(X,\xi _{n})$).

So that $g\cdot i\cdot F\delta _{n}=g\cdot \delta _{n+1}$ but also $g\cdot i\cdot F\delta _{n}=x\cdot Fg\cdot F\delta _{n}=x\cdot F\xi _{n}=\xi _{n+1}$

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You start with an arbitrary diagram, which can't be right. Perhaps you should start with the initial object (of $\mathcal{C}$). –  Zhen Lin Jun 26 at 7:41
    
@Zhen: I started with a diagram of the form: $0\rightarrow F0\rightarrow F^{2}0\rightarrow \cdot \cdot \cdot \cdot \rightarrow F^{n}0$. I'm pretty sure the first part is right, I just need uniquenes. –  Chilango Jun 26 at 13:46
    
OK, so you have a homomorphism of $F$-algebras $f : A \to X$. Compose it with the various colimit cocone components $F^n 0 \to A$. What can you say? –  Zhen Lin Jun 26 at 13:49
    
I am using the cocone $(FA,\delta _{n}), n>0$ and have to show that if $(X,\xi _{n})$ is the cocone for $X$, then $f\cdot \delta _{n}=\xi _{n}$. Of course, $x\cdot Ff=f\cdot i$, and $i$ is an isomorphism. So I have two cones (two paths from $FA$ to $X$ and I want to compare them. I know there is a way to simplify the diagram chase by defining a new category whose objects are diagrams and whose arrows are pairs of functors and natural transformations. But I want to do it just using the cocones. –  Chilango Jun 26 at 14:09
    
Proceed by induction. Of course, $f \circ \delta_0 = \xi_0$ because $0$ is initial. The induction step uses the fact that $F$ preserves $\omega$-colimits. –  Zhen Lin Jun 26 at 14:27

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