Directed limits of topological spaces and embeddings Let $(\{X_\alpha\}_\alpha,\{f_{\alpha\beta}:X_\alpha\rightarrow X_\beta\}_{\alpha\preceq\beta})$ be a directed system of topological spaces and $(X,\{g_\alpha:X_\alpha\rightarrow X\}_\alpha)$ its direct limit.  Can we conclude that the $g_\alpha$ are embeddings if the $f_{\alpha\beta}$ are?
I think I can prove it for the case that $({\bf N},\leq)$ is cofinal in the directed set...
 A: So let $(A,\le)$ be a directed set and let $(X_\alpha)_\alpha$ be a directed system over $A$, that is, $f_{αβ}:X_α\to X_β$ is a map whenever $α\le β$, and for $α\leβ\leγ$ we have $f_{αγ}=f_{βγ}\circ f_{αβ}$. In particular, if $α\leβ$ and $β\leα$, then $f_{αβ}\circ f_{βα}=f_{ββ}=\mathbf 1_{X_β}$. The colimit $X$ of this diagram is the space
$$X=\left(\coprod X_α\right)/(x\sim f_{αβ}(x))$$
It is easy to show by induction that two points $x\in X_α$ and $y\in X_β$ are identified if and only if there is a $γ\in A$ such that $f_{αγ}(x)=f_{βγ}(y)$.
If all these maps are injective, and $α=β$, then for two identified points $x,y$ we have $x=y$, so the map $g_α:X_α\to X$ is injective.
If, moreover, all $f_{αβ}$ are embeddings, then the question arises whether the $g_α$ are embeddings as well. This means just that the quotient map $g:\coprod X_α\to X$ restricts to a quotient map $g_α:X_α\to g(X_α)$ on each $X_α$, which is the case if for every closed set $C_α\subseteq X_α$, there is a closed and $g$-saturated set $C\subseteq \coprod X_α$ such that $C_α=C\cap X_α$.
Let us assume for the moment that all embeddings are closed. Let $C_α$ be closed in $X_α$. For every $γ\geα$, put $C_γ=f_{αγ}(C_α)$, and for every $β\in A$, let $C_β=f_{βγ}^{-1}(C_γ)$ for some $γ\geα,β$. This is independent on the choice of $γ$, for if $δ\geα,β$ and $f_{βγ}(y)=f_{αγ}(x)$, then for some $ϵ\geγ,δ$ we have 
$$
f_{δϵ}(f_{βδ}(y)) = f_{βϵ}(y) = f_{γϵ}(f_{βγ}(y)) = f_{γϵ}(f_{αγ}((x)) = f_{αϵ}(x) = f_{δϵ}(f_{αδ}(x))
$$
and thus $f_{βδ}(y)=f_{αδ}(x)$. Let $C=\coprod C_β$. Then


*

*$C$ is just the saturation of $C_α$ since it contains all the points which have a common image with some point in $C_α$, and no point apart from these. This also shows that $C\cap X_α=C_α$.

*$C$ is closed since all $C_γ,γ\geα$, are closed since the maps $f_{αγ}$ are closed embeddings, and the remaining $C_β$ are preimages of closed sets.


This shows that each $g_α$ is a homeomorphism onto its image. In fact, it is a closed embedding
I have to admit that I'm not sure about the case where the embeddings are not closed. Let me think about that for a while.
Edit: Okay, it seems like the embeddings don't have to be closed. Given a closed set $C_α\subseteq X_α$, we can set $C_γ=\overline{f_{αγ}(C_α)}$ for each $γ\geα$, and $C_β=f_{βγ}^{-1}(C_γ)$ for each $β\in A$. To show independence of the choice of $γ$, let $δ$ be another index larger than $α$, and choose some $ϵ\geγ,δ$. Assume that $f_{βγ}(y)\in{C_γ}$ and let $U$ be an open neighborhood of $f_{βδ}(y)$. Then there is an open set $V$ such that $V\cap f_{δϵ}(X_δ)=f_{δϵ}(U)$ and this $V$ contains $f_{βϵ}(y)$. It follows that $f_{γϵ}^{-1}(V)$ is an open neighborhood of $f_{βγ}(y)$ and thus contains some $f_{αγ}(x)$. But then $f_{αϵ}(x)\in V$. which implies that $f_{αδ}(x)\in U$. Hence $f_{βδ}(y)\in C_δ$.
So let $C=\coprod C_β$. Then


*

*$C\cap X_α=C_α$

*$C$ is closed by definition and by continuity of the maps

*$C$ is saturated as one may readily verify


This proves that $g_α$ is an embedding.
