Let $f:X\to Y$ be a continuous epimorphism where $X$ is connected. Show that $Y$ is connected.
Now I know the concept of epimorphism and connectedness, just not sure how those two are linked with each other. I'm thinking of proving by contradiction, but not sure how to proceed from there since epimorphism concerns $X \to Y \rightrightarrows Z$?
In the category of topological spaces, an epimorphism is just a surjective continuous map. It is enough to show the continuous image of a connected set is connected. Then since the morphism is surjective, it will follow that $Y$ is connected.
To prove this, suppose that the image was disconnected. Then we could partition $Y$ into two sets $A,B$ which are both open and closed, disjoint and non-empty. However, since the morphism (which we'll name $f$) is continuous, we could then write $X = f^{-1}(A) \cup f^{-1}(B)$. Since $A$ and $B$ are disjoint, so are their pre-images. These are open sets, and since they are each other's complements, they are closed. They are non-empty since the map is surjective.
This partitions $X$, and this was supposed to be impossible by connectedness. This contradiction proves $Y$ is connected.