If $f$ is continuous, then $G$ is connected . True/false?

Question

Let $X$ be a compact topological space and let $f : X \rightarrow \mathbb{R}$ be a function . The graph $f$ is the set $G = \{ (x, f(x) ) : x \in X \} \subseteq X \times \mathbb{R}$

My question is that Is the following statement is True/false ?

If $f$ is continuous, then $G$ is connected

My attempt : I think yes, because the continuous image of a connected set is connected.

Answer 1

In general, it is false. If, for instance, $X=\{0,1\}$, endowed with the discrete topology, then $X$ is compact, but $G$ is always disconnected.

Of course, the statement holds if $X$ is connected, even without assuming compactness.

Answer 2

False, and compactness is a red herring. We need $X$ to be connected.

It's clear that $G$ is homeomorhic to $X$ when $f$ is continuous (the first projection is the continuous inverse and when $f$ is continuous, so is the map $F: x \to (x, f(x))$ and $G=F[X]$).