Let f : X → Y be a continuous surjective function.
(i)If X is sequentially compact then Y is sequentially compact.
(ii) If X is compact then Y is compact.
(i)If X is sequentially compact then Y is sequentially compact.
(ii) If X is compact then Y is compact.
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Proof summaries
i) Suppose you have an infinite sequence in Y. Since f is surjective, you can choose a sequence in X so that each element of that sequence is mapped to the corresponding member of the original sequence. Since X is sequentially compact, every infinite sequence has a convergent subsequence. Take any such convergent subsequence and map it back to Y. Since f is continuous, it maps convergent sequences to convergent sequences. Thus, we have found a convergent subsequence of the original sequence.
ii) Take an open cover "O" of Y. For an open set U in O, f^-1(U) is open of X. Thus, we can pull the open cover back into X to find an open cover of X. Since X is compact, we can find a finite sub-cover, S. We can map each f^-1(U) in S back to a U in Y. Since S covers X and f is surjective, f(S) covers Y. Thus, f(S) is a finite subcover of O.
i) Suppose you have an infinite sequence in Y. Since f is surjective, you can choose a sequence in X so that each element of that sequence is mapped to the corresponding member of the original sequence. Since X is sequentially compact, every infinite sequence has a convergent subsequence. Take any such convergent subsequence and map it back to Y. Since f is continuous, it maps convergent sequences to convergent sequences. Thus, we have found a convergent subsequence of the original sequence.
ii) Take an open cover "O" of Y. For an open set U in O, f^-1(U) is open of X. Thus, we can pull the open cover back into X to find an open cover of X. Since X is compact, we can find a finite sub-cover, S. We can map each f^-1(U) in S back to a U in Y. Since S covers X and f is surjective, f(S) covers Y. Thus, f(S) is a finite subcover of O.