# Show that inverse of this continuous function is continuous

This question is from baby Rudin, Ch 5, exer 2:

Suppose $$f'(x)>0$$ in $$(a,b)$$. Hence $$f$$ is strictly increasing in $$(a,b)$$. Let $$g=f^{-1}$$. Prove $$g'(f(x))=\frac1{f'(x)}$$.

The solution I have uses the fact thaty $$g$$, the inverse of $$f$$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $$f$$ is continuous bijection from compact metric space $$X$$ to metric space $$Y$$, then $$f^{-1}$$ continuous from $$Y$$ to $$X$$.'

Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $$[c,d]\subset(a,b)$$ and noting $$g|_{f([c,d])}:f([c,d])\to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.

So, how to show that $$g$$ indeed continuous on $$(a,b)$$?

• Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x \in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25, 2018 at 13:58
• As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25, 2018 at 14:05
• @xbh, thanks for reply. But, if we take function $f$ from $[0,1)\cup[2,3]\to[0,2]$ defined by $f(x)=x$ if $x\in[0,1)$ and $f(x)=x-1$ if $x\in[2,3]$, and consider $[c,d]\subset[0,1)$ and $[c,d]\subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not! Nov 25, 2018 at 14:18
• @Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)\cup[2,3])$ and that is all you need. Nov 25, 2018 at 14:44

It is sufficient to show that $$f$$ is an open map. So for any $$(c,d)\subset (a,b)$$, observe that $$f((c,d))=(f(c),f(d))$$ since $$f$$ is a stricly increasing function. Hence $$f$$ maps any open set into an open set.