If the composition $f\circ g$ of $f$ and $g$ is well defined and is continuous, does that necessarily imply that $f$ is continuous and $g$ is continuous?


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Counterexample: $f$= every discontinuous bijection, $g=f^{-1}$

  • $\begingroup$ Some more surprising counterexample? That almost works? $\endgroup$ – kjetil b halvorsen Feb 10 at 11:43
  • $\begingroup$ How do you define "almost"? $\endgroup$ – YuiTo Cheng Feb 10 at 11:45
  • $\begingroup$ Maybe something which is continuous in all but a few points? $\endgroup$ – kjetil b halvorsen Feb 10 at 11:57
  • $\begingroup$ Then you just plug in your function as $f$ and its inverse as $g$, as long as it's a bijection. $\endgroup$ – YuiTo Cheng Feb 10 at 11:59

No. Let$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x=0\\0&\text{ otherwise}\end{cases}\end{array}$$and let $g\colon\mathbb{R}\longrightarrow\mathbb R$ by any discontinuous function such that $0$ doesn't belong to its range. Then $f\circ g$ is continuous (it is constant, actually), but both $f$ and $g$ are discontinuous.


If you use the Dirichlet function for both $f$ and $g$:

$f \circ g$ is continuous (it's actually constant, $x \rightarrow 1$), but both $f$ and $g$ are nowhere continuous.


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