# Pitfalls of open maps

Suppose you have an open map $$p$$ between topological spaces, and if you have a subet $$A$$ of $$p$$’s domain such that $$p(A)$$ is open. Can you then conclude that $$A$$ is open? Nope! Consider the following spaces $$X=\{x_1,x_2\}$$ and $$Y=\{y_1,y_2\}$$ with topologies $$\tau_X=\{\varnothing, X, \{x_1\}\}$$ and $$\tau_Y=\{\varnothing,Y,\{y_1\}\}$$, respectively and let $$p: X\times Y\to X$$ be the projection onto its first fator. This is an open map. If we consider $$A=X\times\{y_2\}$$ we see that $$A$$ is not open in $$X\times Y$$, but we have that $$p(A)=p(X\times\{y_2\})= X$$ which is trivially open in $$X$$.

# Is this quotient space connected?

I came across this little problem recently: If $$X$$ is a topological space with exactly two components, and given an equivalence relation $$\sim$$ what can we say about its quotient space $$X/{\sim}$$? It turns out that $$X/{\sim}$$ is connected if and only if there exists $$x,y\in X$$ where $$x$$ and $$y$$ are in separate components, such that $$x\sim y$$.

Suppose first that there exists $$x,y\in X$$ such that $$x\sim y$$. Let $$C_1$$ and $$C_2$$ be the two components of $$X$$ and let $$p: X \to X/{\sim}$$ be the natural projection. Since $$p$$ is a quotient map it is surely continuous and since the image of a connected space under a continuous function is connected we have have that, say $$p(C_1)$$ is connected and so is $$p(C_2)$$, but since $$x\sim y$$ we have that $$p(C_1)\cap p(C_2)\neq \varnothing$$ so $$X/{\sim}$$ consists of a single component, becuase $p(C_1)\cup p(C_2) = p(C_1\cup C_2)=p(X)=X/{\sim},$ as wanted.

To show the reverse implication, we use the contrapositive of the statement and show: if we for no $$x\in C_1$$ or $$y\in C_2$$ have that $$x\sim y$$, then $$X/{\sim}$$ is not connected. Assume the hypothesis and note that then $$p(C_1)$$ and $$p(C_2)$$ are then disjoint connected subspaces whose union equal all of $$X/{\sim}$$ (since $$p$$ is surjective). But then the images of $$C_1$$ and $$C_2$$ under $$p$$ are two components of $$X/{\sim}$$, showing that $$X/{\sim}$$ is not connected. As wanted.