 Solution

ID: past-exam-of-the-topology-course-of-the-university-of-oxford/2015/2/a/ii/solution

Let

(V_{i})

be an open cover of

Y

f^{- 1} (V_{i})

are open in

X

; by surjectivity, they cover

X

. Compactness supplies finitely many covering

X

. Applying

f

and using

f (X) = Y

shows that the corresponding finitely many

V_{i}

cover

Y

Thus a continuous surjective function from a compact space has compact target. This is the continuous image of a compact space theorem.

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