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 $$A | B$$ is a partition of $$T_2$$.

Then it follows that $$f^{-1} \left({A}\right) | f^{-1} \left({B}\right)$$ 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.