Continuous Image of Connected Space is Connected/Proof 1

Proof
Let $i: f \left({T_1}\right) \to T_2$ be the inclusion mapping.

Let $g: T_1 \to f \left({T_1}\right)$ be the surjective restriction of $f$.

Then $f = i \circ g$.

Hence, by Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is continuous.

We use a Proof by Contradiction.

Suppose that $A \mid B$ is a partition of $f \left({T_1}\right)$.

Then it follows that $g^{-1} \left({A}\right) \mid g^{-1} \left({B}\right)$ is a partition of $T_1$.