Prove That the Inverse Map is Continuous
Proving that the inverse of a bijective continuous function is continuous
Solution 1
I believe the relevant proof in the abstract topological case is that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism as can be seen here.
However regarding the proof, $B$ has to be an interval since again abstractly the continuous image of a connected set is connected, in other words intervals are mapped to intervals. It can also be shown that $B$ needs to be a closed interval, topologically you could consider compactness.
Continuous and bijective implying monotonic is using a property of the interval.
Notice that $f$ and $f^{-1}$ strictly monotonically increasing, guarantees $y_+>y_-$. Now if $\delta_\epsilon=\min\{y_+-y_0,y_0-y_-\}$ which are both positive numbers, then supposing $\delta_\epsilon=y_+-y_0$, then $|y-y_0|<\delta_\epsilon$ means $y\in (y_0-\delta_\epsilon,y_0+\delta_\epsilon)=(2y_0-y_+,y_+)\subset (y_-,y_+)$ since if we subtract the bigger number $y_0-y_-$ from $y_0$, we get $y_-$. It's similar the other way round.
I also feel like the last step is a little brief, but essentially $y\in (y_-,y_+)$ and using monotonicity guarantees $$f^{-1}(y_0)-\epsilon=x_0-\epsilon=f^{-1}(y_-)\leq f^{-1}(y)\leq f^{-1}(y_+)=x_0+\epsilon=f^{-1}(y_0)+\epsilon$$
Solution 2
Your question seem to imply that $B$ is a subset of the real numbers (this statment stays true if $B$ is any Hausdorff topological space). A function $g$ is continuous if $g^{-1}(A)$ is closed for every closed set $A$. Then to show that $f^{-1}$ is continuous, we have to show that the image of every closed set is closed.
Let $A$ be a closed subset of $[a,b]$, then $A$ is compact since it is the intersection of a closed set ($A$) and a compact set ($[a,b]$). Since compactness is a topological invariant, $f(A)$ must be compact which means that $f(A)$ is closed.
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Comments
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I'm trying to prove the following statement: If $f: [a,b] \rightarrow B$, where $B$ is a subset of the reals, is continuous and bijective then it's inverse $f^{-1}$ is continuous.
Firstly we note that $f^{-1}$ is monotonic since $f$ is monotonic. Say we have $f^{-1}$ is strictly increasing (the proof for it being strictly decreasing would be similar), then I have the following proof for its continuity:
I can understand everything in the proof except from the chosen value for $\delta_{\epsilon}$. I think $\delta_{\epsilon}$ should be chosen to be $\delta_{\epsilon}=max\{y_+ - y_0 , y_0- y_-\}$ because if we had $\delta_{\epsilon}=min\{y_+ - y_0 , y_0- y_-\}$ and we had that $y_+ - y_0>y_0 - y_-$ then $\delta_{\epsilon}=y_0 - y_-$ and $|y-y_0|<\delta_{\epsilon} \rightarrow y \in (y_-, y_0)$ not $y \in (y_0, y_+)$ so we also wouldn't have $y \in (y_-,y_+)$. However, I'm not sure if I'm right or not and if the proof is actually right.
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What is $B$ here?
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@B.Pasternak $B$ is just some range.
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Any valid proof is going to need to use something special about $[a,b]$, or possibly about $B$ if it's specified what that is. It's not in general true that if a map between two topological spaces (or even two metric spaces) is continuous and bijective then its inverse is continuous. For a counterexample, consider the map $[0, 2 \pi)$ to the unit circle $S^1 = \{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1 \}$ given by $t \mapsto (\cos t, \sin t)$.
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@DanielSchepler Your map is not a counterexample since $[0,2\pi)$ is not a closed interval.
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I should also probably say that $B$ is a subset of the reals.
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@T.Haddad It is when considered as a topological space in its own right.
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@B.Pasternak You are right, $[0,2\pi)$ is closed in $[0,2\pi)$. But in his question, the domain was about a closed interval of the real numbers.
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@T.Haddad I know, I was just making the comment that the proof as given doesn't seem to be using enough about the domain for me to be convinced it would fail on this counterexample. So there must be something missing in the argument.
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@DanielSchepler The full proof used the fact that $f$ is monotonic but I didn't include in the picture but I mentioned it instead. Maybe that's what's missing?
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@T.Haddad I also don't doubt that you know, but since this questions concerns details, I'd thought I'd mention it, may it be of use to the OP.
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OK, at the time I posted the comment I didn't know $B$ was necessarily a subset of the reals.
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I feel like this is using "a compact subset of a Hausdorff space is closed" without explicitly saying it.
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How do we know that $2y_0 - y_+ < y_+$?
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@ReinhildVanRosenú Oh that's because $\delta_\epsilon$ is a positive number and $y_0-\delta_\epsilon<y_0+\delta_\epsilon$, and in the case I mentioned above just sub in $y_+-y_0$ for $\delta_\epsilon$.
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@ReinhildVanRosenú it's the same reasoning the other way round. If $\delta_\epsilon=y_0-y_-$ then $|y-y_0|<\delta_\epsilon$ is equivalent to $y\in (y_0-\delta_\epsilon,y_0+\delta_\epsilon)=(y_-,2y_0-y_-)$, just by subbing in, and again since $\delta_\epsilon\leq y_+-y_0$, we can see that $(y_0-\delta_\epsilon,y_0+\delta_\epsilon)\subset (y_0-\delta_\epsilon,y_0+y_+-y_0)$, or by subbing in for $\delta_\epsilon$ and simplifying, $(y_-,2y_0-y_-)\subset (y_-,y_+)$.
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Source: https://9to5science.com/proving-that-the-inverse-of-a-bijective-continuous-function-is-continuous
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