Continuous Image of Compact Space is Compact/Proof 1

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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 compact then so is its image $f \sqbrk {T_1}$ under $f$.


That is, compactness is a continuous invariant.


Proof

From Restriction of Mapping to Image is Surjection, the restriction of $f$ to $\Img f$ is a surjection.

The result follows from Compactness is Preserved under Continuous Surjection.

$\blacksquare$