Let $X$ be a compact topological space, $f_i:Y_i\to X$ a family of continuous maps such that the topology on $X$ is final for it (i.e., $U\subset X$ is open iff $f_i^{-1}(U)$ is open for each $i$, for more categorical formulation see http://en.wikipedia.org/wiki/Final_topology).
Does there exist a finite subfamily with the same property?
Compactness assumption is necessary here, because any open cover of $X$ gives rise to a family of inclusions, for which the topology on $X$ is final. So, the question above can be reformulated as follows: is it true that a topological space $X$ is compact iff in any family $f_i:Y_i\to X$ which generates the topology on $X$ as the final topology one can find a finite subfamily with the same property.
Thank you for fast answers! I am very impressed. The negative answer means that open coverings are very far from families of maps defining topology. Perhaps, open covering are maps from rather than to the space, like locally finite coverings. My real motivation was to understand compactness in terms of the category of topological spaces. I cannot shape this as a well-posed problem, but nonetheless still hope for a reasonable answer.
Best Answer
I think the answer is no. For an example, let $X$ be the unit interval $[0,1]$ (with the usual topology).
For each $i \in \mathbb{N}^+$, let $Y_i = (1/i,1]$ with the usual topology. The obvious inclusion $f_i \colon Y_i \to X$ is continuous.
Let $Y_0$ be $[0,1]$ with an unusual topology: a set is open in $Y_0$ if and only if either it does not contain $0$ or it contains a neighborhood of $0$ in the usual topology. Any subset of $[0,1]$ that is open in the usual topology is open in $Y_0$, so the identity map $f_0 \colon Y_0 \to X$ is continuous.
It is obvious that no finite subfamily is final, because the union of a finite subset of $Y_1, Y_2, \ldots$ must miss some neighborhood $[0,\epsilon)$ of $0$. So every subset of $[0,\epsilon)$ that does not contain $0$ is open in the final topology for this subfamily.
On the other hand, the topology on $X$ is final for the entire family: suppose $f_i^{-1}(U)$ is open for every $i$.
If $0 \notin U$, then $U = \bigcup_{i=1}^\infty (U \cap Y_i)$ is the union of open sets, and is therefore open.
If $0 \in U$, then, since $U$ is open in the topology of $Y_0$, it must contain a neighborhood $[0,1/n)$ of $0$ in the usual topology. Then $U = [0,1/n) \cup ( U \cap Y_{2n})$ is the union of two open sets, and is therefore open.