Suppose that $X$ is a space filtered by closed subspaces $X\_{1}\subset X\_{2}\subset \dots$. 


As topological space $X=\operatorname{colim}\_{n}X\_{n}$.
We define $Y\_{n}=X\_{n+1}/X\_{n}$, and consider the induced maps $Y\_{n}\rightarrow Y\_{n+1}$. Let define $Y=\operatorname{colim}\_{n}Y\_{n}$. 


*Question:* $X$ is homeomorphic to $Y$?



Filtered colimit of a topological space
As far as I can tell, this is not true in general.


Consider simply $X\_n=X$. In this case $Y\_n$ (and consequently $Y$) will be a singleton.


