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Could you tell me why theorem 4.17 does not just follow from theorem 4.8?

I mean inverse mapping $f^{-1}$ will be continuous on $Y$ iff $f(V)$ is open in $Y$ for every open $V$ in $X$ by theorem 4.8, and the statement, "$f(V)$ is open in $Y$ for every open $V$ in $X$" is true since $f$ is continuous. Hence $f^{-1 }$ will be continuous on $Y$.

Why is proof for 4.17 in the textbook so long?


$\mathbf{4.17}\,\,\,$*Theorem*$\quad$*Suppose $f$ is a continuous 1-1 mapping of a compact metric space $X$ onto a metric space $Y$. Then the inverse mapping $f^ {-1}$ defined on $Y$ by $$f^ {-1}(f(x))=x\quad(x\in X)$$is a continuous mapping of $Y$ onto $X$.*

*Proof*$\quad$Applying Theorem $4.8$ to $f^{-1}$ in place of $f$, we see that it suffices to prove that $f(V)$ is an open set in $Y$ for every open set $V$ in $X$. Fix such a set $V$.

The complement $V^c$ of $V$ is closed in $X$, hence compact (Theorem 2.35); hence $f(V^c)$ is a compact subset of $Y$ (Theorem 4.14) and so is closed in $Y$ (Theorem 2.34). Since $f$ is one-to-one and onto, $f(V)$ is the complement of $f(V^c)$. Hence $f(V)$ is open.

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$\mathbf{4.8}\,\,\,$*Theorem*$\quad$*A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous on $X$ if an only of $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$.*


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Your assertion "$f(V)$ is open for every open $V$ in $X$ since $f$ is continuous" is not true. Hence you cannot use Theorem 4.8 to conclude.

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  • $\begingroup$ Examples: $f$ could send everything to a point. Then $f$ would be continous, but not 1-1, $f^-1$ would not be a function from $Y$ to $X$. Or $f: \mathbb{R} \rightarrow \mathbb{R}^2$ could send the number line to a line in the plane. The number line is open, but a line in the plane is closed. Here the problem is that our $X$ is not compact. $\endgroup$
    – Greebo
    Commented Aug 7, 2013 at 7:47

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