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Let $f : M \mapsto N$ be continuous. Consider a compact $K \subseteq M$.

Assume as given: $\forall B \subseteq N$ open, $f^{-1}(B)$ is open in $M$.

  1. Consider an open cover of $f(K)$, call it $A_i$ for $i \in I$.
  2. Then, $B_i$ is an open cover of $K$, where $B_i = \{x \in M: f(x) \in A_i \}$
  3. There exists a finite subcover of $K$, call it $B_{n_1}, ..., B_{n_k}$
  4. Then, $A_{n_1}, ..., A_{n_k}$ is a finite subcover of $f(K)$

Just need to verify this proof, in particular, I am concerned if $(2)$ and $(4)$ need more details.

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  1. What do you mean by calling the open cover $A_i$? I would start 'Let $U$ be an open cover of $f(K)$.'
  2. Yes, more details would be welcome. Why is what you denote $B_i$ an open cover of $K$? Answer: for every $x \in K$, $f(x)$ belongs to some $u \in U$.
  3. Fine.
  4. Again, more details would be helpful. Answer: for every $y \in f(K)$ we have $y = f(x)$ where $x \in K$. Then $x \in f^{-1}(U_i)$ for some $i$ and so $y = f(x) \in U_i$.

A very good start to the proof nonetheless. If I were grading I would probably award most of the points, but not all, for the omission of the above.

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It seems good to me.

(2) works out because $B_i = f^{-1}(A_i)$ is open for all $i$ since $f$ is continuous, and certainly covers $K$, so it is indeed an open cover as you stated.

In a similar fashion for statement (4), $A_{n_1}, \dots, A_{n_k}$ is a finite sub cover since for any $y \in f(K), \; \exists i : x \in B_{n_i} $ (Since $\{B\}_{i \in I}$ is an open cover for $K$) so that $ f(x) =y \in A_{n_i}$ by construction.

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