This is a continuation of an original question about spreads, which are something like pre-branched covering spaces. See the basic definitions here: A Complete Spread
I have an example of a spread which is not complete (coming from Montesinos, Covering Spaces After Fox), but I don't even understand why it's a spread.
Example: $g:Y\to Z$ where $$Y=\{z\in\mathbb{C}||z|=1\}-\{i\},\quad Z=\mathbb{R},\quad g(z)=Re(z).$$ Allegedly, this is a spread where $Y,Z$ are metrizable, connected, and locally connected, but not complete. But I don't understand how this can be a spread, considering what happens to the point $1\in\mathbb{C}$.
The inverse images of open sets in spreads are supposed to form a base for the topology of the antecendents $Y$. So $g(1)=1 \in Z$, but an open set in $\mathbb{R}$ which contains 1 would be $(1-\epsilon, 1+\epsilon)$, and the inverse image of this does not lie in $Y$.
Can someone help me understand how this can be a spread?
Following the helpful comment from Daniel:
The map $g:Y\to Z$ is a spread because every preimage of $g$ is an element of the topology of $Y$. For the point $1\in\mathbb{R}$, consider the image of an open set in $Y$:
$$g((e^{-i\epsilon},e^{i\epsilon}))=(-\cos\epsilon,1)$$
Since the preimage contains only elements which are in the domain of the function, the preimage of a set $(1-\delta,1+\delta)$ is a set in $Y$ with appropriate values of $\epsilon$.
It is not complete because completeness has to occur on each connected component separately. Consider $0\in\mathbb{R}$. The component around $-i\in\mathbb{C}$ has one point in the intersection of all preimages of the open neighborhoods of $0$. However, the component around $i$ (but not including $i$) does not satify this. The intersection of all the preimages of the neighborhoods of $0$ is empty.