Continuous Image of Connected Space is Connected

Theorem
Let $T_1$ and $T_2$ be topological spaces.

Let $f: T_1 \to T_2$ be a continuous mapping.

If $T_1$ is connected, then so is $f \left({T_1}\right)$.

Corollary 1
Connectedness is a topological property.

Corollary 2
Let $T$ be a connected space.

If $f: T \to \R$ is a continuous mapping then $f \left({T}\right)$ is an interval.

Corollary 3
If $f: \left[{a. . b}\right] \to \R$ is continuous then $f$ has the intermediate value property.

Proof
By Continuity of Composite with Inclusion, the surjective restriction $f_1: T_1 \to f \left({T_1}\right)$ induced by $f$ is continuous when $f$ is continuous.

So it is enough to prove the result where $f: T_1 \to T_2$ is a surjection.

So, in this case, suppose $A \mid B$ is a partition of $T_2$.

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

Hence the result.

Proof of Corollary 1
Follows directly from the above and definition of topological property.

Proof of Corollary 2
From the main result, the continuous image of the connected space $T$ is connected.

The result follows from Only Intervals are Connected.

Proof of Corollary 3
Since $\left[{a. . b}\right]$ is connected, then by corollary 2 so is $f \left({\left[{a . . b}\right]}\right)$ and hence $f \left({\left[{a . . b}\right]}\right)$ is an interval.

The result follows by definition of an interval and the I.V.P..