Is the quotient map $q:X \times I \to CX$, where $CX$ is the cone of $X$, is open or closed?

Question

Suppose $CX$ is the cone of $X$ defined by $CX := X \times I/ X \times \{0\}$.

My question is the corresponding quotient map $q:X \times I \to CX$ is open or closed.

Intuitively, it seems this should be true if I picked a nice space $X$. But there could be a pathological space where this could fail. If it does, is there a condition on the space $X$ which can ensure that the quotient map is open or closed?

I was thinking about this question because I was reading about different construction of join of spaces and trying to see how they are related to each other.

Answer 1

If $X$ is compact and Hausdorff, then the quotient map is closed for free. Indeed $X \times I$ is compact and $CX$ is Hausdorff (not difficult to check what happens when you quotient out by a closed subset)

In this case the projection is not open in general. For example just take $X = S^1$, and let $J \subset X$ an open connected subset. Then $$q^{-1}(q(J \times I)) = (J \times I) \cup (X \times \{0\})$$ which is not open. As $CX$ has the quotient topology, then $q(J \times I)$ is not open in $CX$.

Answer 2

Consider a space $X$ with subspace $A$, and the quotient map $q\colon X\to X/A$ contracting $A$ to a point. If $A$ is open (or closed) this map is open (or closed).

For the open case, let $A$ be open and consider an open subset $B$. The set $q^{-1}q(B)$ is either $B$ or $A\cup B$, so necessarily open. For the closed case, let $A$ be closed and consider a closed subset $B$. The set $q^{-1}(q(B)^{\mathrm{c}})$ is either $B^\mathrm{c}$ or $B^\mathrm{c}\setminus A$, which are both open sets, as $B$ and $A$ are closed. This implies $q(B)^\mathrm{c}$ is open, so $q(B)$ is closed.

Since the subset $X\times \{0\}$ is closed in $X\times I$, the map to the cone is closed.