Let \(Z=X\cup Y\), for \(X\) and \(Y\) connected subspaces of \(Z\) with \(X\cap Y=\varnothing\). Let \(x_0\in X\) and \(y_0\in Y\). Let \(\sim\) be the equivalence relation generated by the equivalence \(x_0\sim y_0\). Show that the quotient space \(Z/\sim\) is connected.
Solution
Let \[ q:Z\longrightarrow Z/\!\sim \] be the quotient map.
1. The subsets \(q(X)\) and \(q(Y)\) of \(Z/\!\sim\) are connected.
The restrictions \[ q|_X:X\longrightarrow q(X), \qquad q|_Y:Y\longrightarrow q(Y) \] are continuous and surjective.
To see directly that a continuous image of a connected space is connected, suppose for instance that \(q(X)\) had a separation \[ q(X)=U\cup V \] into disjoint nonempty sets open in the subspace topology. Then \[ (q|_X)^{-1}(U), \qquad (q|_X)^{-1}(V) \] would be disjoint nonempty open subsets of \(X\) whose union is \(X\), contradicting connectedness of \(X\). Thus \(q(X)\) is connected, and the same argument gives connectedness of \(q(Y)\).
2. The sets \(q(X)\) and \(q(Y)\) have nonempty intersection.
The equivalence relation identifies \(x_0\) and \(y_0\), so \[ q(x_0)=q(y_0). \] Since \(x_0\in X\) and \(y_0\in Y\), this common point lies in \[ q(X)\cap q(Y). \]
3. If two connected subsets of a space have nonempty intersection, then their union is connected.
Let \(C,D\) be connected and suppose \(C\cap D\ne\varnothing\). If \(C\cup D\) admitted a separation into disjoint nonempty relatively open sets \(U,V\), connectedness of \(C\) would force \(C\) to lie entirely in one of \(U,V\), and connectedness of \(D\) would force \(D\) to lie entirely in one of them. Because \(C\cap D\ne\varnothing\), they cannot lie in different members of the separation. Hence both \(C\) and \(D\) lie in the same member, leaving the other empty, a contradiction. Therefore \(C\cup D\) is connected.
4. The quotient space \(Z/\!\sim\) is connected.
Because \(Z=X\cup Y\) and \(q\) is surjective, \[ Z/\!\sim=q(Z)=q(X)\cup q(Y). \] By 1 these two subsets are connected, and by 2 they intersect. Applying 3 proves that their union, hence \(Z/\!\sim\), is connected.