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. 



Possible categorical reformulation for the usual definition of compactness
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.


