Connected components of quotient space.

Question

Let $\mathcal{R}$ be an equivalent relation over $X$ such that $\forall x\in X$, its class $[x]$ is included in some connected component of $X$. Show that the connected components of $X/\mathcal{R}$ are the images of the connected components of $X$ via the quotient map.

So far, denoting $\pi : X\to X/\mathcal{R}$ the quotient map, I've proved that, given a connected component $C_x$ of $X$, then $\pi(C_x)\subset C_{[x]}$. This is because $x\in [x]$ and the continuous image of a connected set is also connected. But I don't see the other inclusion, could you please tell me any hint?

Answer 1

If $q:X\to Y$ is a quotient map with the property that the fiber of each $y\in Y$ is contained in some component of $X$. Then, it maps each component of $X$ to a component of $Y$. Indeed, if $C$ is a component of $X$, then $q(C)$ is contained in a component $D$ of $Y$. Moreover, $q^{-1}(D)$ is connected. Seeking contradiction, assume $q^{-1}(D)=U\cup V$ where $U$ and $V$ are non-empty disjoint open subsets of $q^{-1}(D)$, then $q(U)$ and $q(V)$ are non-empty disjoint subsets of $D$ with $D=q(U)\cup q(V)$. We claim that $q(U)$ and $q(V)$ are open subsets of $D$. Note that if $x\in q^{-1}(q(U))$, then $q(x)=q(a)$ for some $a\in U$. But, $q(x)=q(a)=y\in D$ (as $q$ is onto) and hence $a,x\in q^{-1}(y)\subset U$ as $q^{-1}(y)$ is contained in a connected subset. Thus, $U=q^{-1}(q(U))$ and similarly $V=q^{-1}(q(V))$. Since $q^{-1}(D)$ is a closed subset, $q:q^{-1}(D)\to q(q^{-1}(D)=D$ is a quotient map which implies that $q(U)$ and $q(V)$ are open, a contradiction.

Now, $q(C)\subset D$ and thus $C\subset q^{-1}(q(C))\subset q^{-1}(D)$. So, $C=q^{-1}(D)$ and hence $q(C)=D$.

Answer 2

It is easier to think about connectedness in terms of the relative topology. You can restrict the domain $X$ to $\pi^{-1}(C_{[x]})$ and assume that the quotient is connected (equal to $C_{[x]}$). Let us show that the domain is also connected.

Take a nonempty clopen set $F \subset X$. Since clopen sets are unions of connected components, the hypothesis imply that $F$ is the union of equivalence classes. But then, this means that $B = \pi(F)$ is clopen. Just notice that $F = \pi^{-1}(B)$. But since the codomain is connected and $F$ is nonempty, $B = C_{[x]}$. But this means that $F = \pi^{-1}(C_{[x]}) = X$. Therefore, $X$ has only trivial clopen subsets. That is, $X$ is connected.

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This proof is a bit difficult because of the first paragraph.

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The quotient topology can be understood as a weaker topology over $X$, where the open sets are restricted to the "saturated ones". That is, the open sets in this weaker topology are the original open sets that are unions of fibers.

When you make the topology weaker, you reduce the number of closed sets and open sets. But the hypothesis warrants that you do not reduce the number of clopen sets!!! Because, in this case, the clopen sets are already unions of equivalence classes.