Showing that the image of a connected space is connected if the function is continuous.

Question

Here is the question:

Show that if $f: X \rightarrow Y$ is any continuous function, and if $X$ is connected, then its image $f(X) \subseteq Y$ (which has the subspace topology) is also connected.

Where we have the following:

we say a space $X$ is connected if the only separations that $X$ has are trivial separations.

A separation of a space $X$ is a continuous function $f: X \rightarrow Z,$ where $Z$ is the discrete two-point space. we say that a separation $f: X \rightarrow Z,$ of the space $X$ is a trivial separation if it is a constant function.

Still I do not know how to prove the question, could anyone help me please?

Answer 1

Suppose $X$ is connected, $f:X\to Y$ is continuous, and $g$ is a continuous function from the image of $f$ to the discrete two-point space $Z$. Then $g\circ f:X\to Z$ is a composition of continuous maps, so continuous, and $X$ is connected, so the composition is constant, so $g$ is constant.

Answer 2

If you can disconnect the image, $f(C)=U\cup V$, with $U$ and $V$ open and disjoint, then $f^{-1}(U)$ and $f^{-1}(V)$ disconnect $C$.

The existence of a separation of $X$ is equivalent to that of a pair of disjoint open sets whose union is $X$.