Let $X\colon I\to\mathsf{Top}$ be a functor, where $I$ is a category and $\mathsf{Top}$ is the category of topological spaces. We then can formulate the limit $\operatorname{lim}_IX$ (with a family of projections $(p_i\colon\operatorname{lim}X\to X(i))_{i\in I}$), which carries the topology generated by $\{p_i^{-1}(U)\mid i\in I, U\subset X(i)\text{ open}\}$. I am wondering if each projection $p_i\colon\operatorname{lim}X\to X(i)$ is an open mapping?
At least this is true for a special case: if we take $I$ to be discrete, then $\operatorname{lim} X=\prod_{i\in I}X(i)$, and projections from a product space to its factors are surjective and open. For the general case the underlying set of $\operatorname{lim}X$ can be identified with a subset $\{(x_i)_{i\in I}\mid X(f)(x_i)=x_j,\forall f\colon i\to j\}$ of $\prod_{i\in I} X(i)$, and its topology can also be identified with the subspace topology, but it seems that $\operatorname{lim}X$ is not always open in $\prod_{i\in I} X(i)$ so whether $p_i$ is open cannot be seen immediately…
Any idea is appreciated.
Best Answer
No, usually they are not. For a really simple example, consider a diagram of the form $$X(0)\stackrel{f}\to X(1).$$ The limit will always be $X(0)$, and the projection to $X(1)$ will be $f$. So, if $f$ is any continuous map that is not open, the projection to $X(1)$ is not open.
Another good example is an equalizer, where we now have two maps $f,g:X(0)\to X(1)$. The limit is then the subspace $E=\{x\in X(0):f(x)=g(x)\}$ and its inclusion map to $X(0)$ will not be an open map unless $E$ happens to be open in $X(0)$.