I asked a similar question yesterday but studying my course further, I stumbled upon a definition of open identification that confused me. It says that open identification is an identification that is also an open map.
The definition of open map that I have states that it is a function that maps open sets to open sets. Now, to my understanding, identification is in fact a continuous, surjective and open map. The definition of open identification does not really make sense then, so I might misunderstood some of the definitions...
Any help would be very much appreciated!
Firstly I'll remark that what you call an "identification" map is actually what one usually calls a "quotient" map.
The following is true:
However, not all quotient maps are open. Let $f : X \to Y$ be a quotient map. The issue is that you may have an open subset of $X$ which is not of the form $f^{-1}(B)$ for any $B \subset Y$.
In fact, call a subset $A \subset X$ saturated if $A = f^{-1}(B)$ for some $B \subset Y$. Then, what is true is that a quotient map takes a saturated open set to an open set.
Lastly, to dispel all doubts, here is an example (from Munkres) of a quotient map that is not an open map:
Let $X = [0, 1] \cup [2, 3]$, and let $Y = [0, 2]$. (Both are given the subspace topology from $\Bbb R$.) Define $f : X \to Y$ by $$f(x) := \begin{cases} x & \text{for } x \in [0, 1], \\ x - 1 & \text{for } x \in [2, 3].\end{cases}$$ I leave it to you to verify that $f$ is a quotient map. However, the image of the open set $[0, 1] \subset X$ is not open in $Y$. Thus, $f$ is not an open map.
(Note that $[0, 1]$ is not saturated since it intersects $f^{-1}(\{1\})$ but does not contain it completely.)