Is there an "opposite" of the inverse limit of topological spaces? Suppose $\{X_i:i\in \omega\}$ is a collection of topological spaces and $f_i:X_i\to X_{i+1}$ is a continuous surjection for each $i\in\omega$.  Is there a space $X$ such that for each $i\in\omega$ there is a continuous  mapping from $X_i$ onto $X$?
If it helps you can assume all spaces are compact Hausdorff.
What I am looking for is sort of the opposite of a inverse limit or product of spaces; inverse limits (with surjective bonding maps) and products map onto each of their factors.
 A: The direct limit works. For $i\in\omega$ let $f_{i,i}=\text{id}_{X_i}$ and $f_{i,i+1}=f_i$. Given $f_{i,j}:X_i\to X_j$ for some $i,j\in\omega$ with $i\le j$, let $f_{i,j+1}=f_{j,j+1}\circ f_{i,j}:X_i\to X_{j+1}$. Let $Y=\bigsqcup_{i\in\omega}X_i$, the disjoint union of the spaces. Define an equivalence relation $\sim$ on $Y$ as follows:

Given $x,y\in Y$, there are unique $i,j\in\omega$ such that $x\in X_i$ and $y\in X_j$; $x\sim y$ iff there is a $k\ge\max\{i,j\}$ such that $f_{i,k}(x)=f_{j,k}(y)$.

Now let $X=Y/\!\sim$, and let $q$ be the quotient map. For each $i\in\omega$ let $q_i=q\upharpoonright X_i$. Finally, give $X$ the final topology with respect to the maps $q_i$: a subset $U$ of $X$ is open in $X$ iff $q_i^{-1}[U]$ is open in $X_i$ for each $i\in\omega$. 
The map $q_i:X_i\to X$ is continuous by construction. Let $x\in X$. There are a $j\in\omega$ and a $y\in X_j$ such that $q_j(y)=x$, and we may assume that $j\ge i$. The bonding maps are surjective, so there is a $z\in X_i$ such that $f_{i,j}(z)=y$, and hence $z\sim y$ and $q_i(z)=x$. Thus, $q_i$ is surjective.
