Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 2 / a

A topological space is connected if it cannot be expressed as a union of two disjoint nonempty sets open in that space. Suppose $f(X)$ had such a separation $U\cup V$. Here $U,V$ are open in the subspace topology of $f(X)$. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$. They are disjoint, nonempty because $U,V$ contain image points, and cover $X$. This contradicts connectedness. Hence a continuous image of a connected space is connected.

A space is path-connected if for every pair $x_0,x_1$ it contains a continuous path $\gamma :[0,1]\to X$ with $\gamma (0)=x_0$ and $\gamma (1)=x_1$. For any $y_0,y_1\in f(X)$ choose preimages $x_0,x_1$. A path between those preimages maps to $f\circ \gamma$, a continuous path from $y_0$ to $y_1$ staying in $f(X)$. Thus a continuous image of a path-connected space is path-connected. Surjectivity onto the ambient codomain $Y$ is not needed; the conclusion concerns $f(X)$.

Suppose a path-connected space had a separation $X=U\cup V$ as above. Choose $u\in U$, $v\in V$, and a continuous path $\gamma$ joining them. The nonempty disjoint sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$ would separate $[0,1]$, contradicting connectedness of an interval. Hence path-connectedness implies connectedness. Equivalently, the connected path image from part (i) cannot meet both sides of a separation.