Let $X,Y$ be connected spaces and $\ f: (X,\mathcal{T}_X) \rightarrow (Z,\mathcal{T}_Z)$, $\ \ $ $g: (Y,\mathcal{T}_Y) \rightarrow (Z,\mathcal{T}_Z)$ be continuous. Construct a quotient space $(X\sqcup Y)/_{\sim}$ by $x\sim y$ if $f(x)=g(y)$.
Show that if $f$ or $g$ is surjective, then $(X\sqcup Y)/_{\sim}$ is connected.
My idea runs as follows:
Assume $f$ is surjective first, and I have a feeling that $(X\sqcup Y)/_{\sim}$ should be homeomorphic to $Z$. Once a homeomorphism, or simply a continuous bijection is defined, then the result is proved.
Define the mapping $$\phi: Z \rightarrow (X\sqcup Y)/_{\sim} $$ by $\phi(z)=[x]$, where $[x]$ is the equivalence class of x such that $f(x)=z$.
It's easy to check that $\phi$ is a well-defined bijective map. But I have no idea how to show the continuity. When I tried to show the pre-image of an open set in the quotient space is open, I noticed that I don't have much information on the open sets in $Z$. So I doubt whether my attempt works.
Any suggestions? Is there any other ways to prove it?
Thank you in advance.
Assume that $f \colon X \to Z$ is onto. Let $i \colon X \to X \sqcup Y$ be the inclusion and $\pi \colon X \sqcup Y \to (X \sqcup Y)/_\sim$ the quotient map. Since $f$ is onto, the composition $\pi \circ i \colon X \to (X \sqcup Y)/_\sim$ is also onto. Hence $(X \sqcup Y)/_\sim$ is connected, as it is a continuous image of a connected space.