Let $I_1,I_2,I_3$ be intervals $\subset \mathbb{R}$. Suppose $f:I_1 \to I_2$ is a surjective continuous function and $g: I_2 \to I_3$ is a discontinuous function. Must the composition $g \circ f$ be discontinuous?

There are some easy counter-examples if $f$ is not assumed to be surjective, e.g. taking $f$ to be a constant function, or in a way that "dodges" the discontinuous point(s) of $g$.

However if such "dodging" is prohibited, I fail to construct such functions nor find an answer from many similar questions on this site. So I am interested to know whether counter-examples exist? If not, is there a proof? Does it have something to do with intermediate value theorem?


3 Answers 3


Here is a proof without compactness of the intervals.

Suppose that $f$ and $g\circ f$ are continuous, $f$ is surjective. I want to show that $g$ is continuous.

Take $x\in I_2$ and a sequence $(x_n)$ in $I_2$ with $x_n\to x$. Since $I_2$ is an interval, there are $x'<x''$ in $I_2$ such that $x\in [x',x'']\subset I_2$ and $x_n\in[x',x'']$ for all $n$. (If this would not be possible, then $I_2$ would be a singleton.)

Since $f$ is surjective, there are $y,y',y''$ in $I_1$ such that $f(y)=x$, $f(y')=x''$, $f(y'')=x''$. Define the interval $J:=[\min(y,y',y''),\ \max(y,y',y'')]$. Due to the intermediate value theorem $f(J) \supset [x',x'']$.

Now for every $n$ there is $y_n\in J$ with $f(y_n)=x_n$. Since $J$ is compact, there is a converging subsequence $(y_{n_k})$ with limit $z$. By continuity of $f$, $f(z)=x$. Then by continuity of $g\circ f$ $$ g(x_{n_k}) = g(f(y_{n_k})) \to g(f(z)) = g(x). $$ Now, we can repeat this argument for each subsequence of $(x_n)$. The limit $g(x)$ does not depend on the chosen subsequence, so $g(x_n)\to g(x)$, and $g$ is continuous.

Can this proof be generalized to higher dimensions?


If $g$ is discontinuous at $x=a$, then $gf$ cannot be continuous. Suppose $gf$ is continuous. For any sequence $\{x_n\}\in I_2, x_n\to a$, due to the surjectivity of $f$, $\exists \{y_n\}\in I_1, f(y_n)=x_n,\forall n$.

$y_n\in I_2$ which is compact, so there must be a subsequesce $y_{n_k}$ that converges. Let $\lim_{k\to \infty}y_{n_k}=L$.

Since $f$ is continuous, $f(L)=\lim_{k\to \infty} f(y_{n_k})=\lim_{k\to \infty}x_{n_k}=\lim x_n=a$. Also, $gf(L)=g(a)$.

So, $\lim g(x_n)=\lim gf(y_{n_k})=gf(L)=g(a)$, thanks to the continuity of $gf$.

The result is that we establish the continuity of $g$ from the continuity of $f$ and $gf$. So $gf$ cannot be continuous.

  • $\begingroup$ $y_n$ has to be in $I_1$. Also compactness of $I_1$ is required. $\endgroup$
    – daw
    May 23, 2019 at 7:47
  • $\begingroup$ @daw But everybody knows it, so no need to write that down.... $\endgroup$
    – Ma Joad
    May 23, 2019 at 10:25
  • $\begingroup$ @daw Ok I will add it. Please tell me anything else that is inaccurate in my answer. $\endgroup$
    – Ma Joad
    May 23, 2019 at 14:38

Given that $g$ is discontinuous there exists $x_0 \in I_2$ and a sequence $(x_n)_{n \in \mathbb{N}} \subseteq I_2$ such that $\lim\limits_{n \to \infty}x_n = x_0$ and $g(x_0) \neq \lim\limits_{n \to \infty}g(x_n)$. Now let $y_0 \in f^{-1}(x_0)$ and for any $n \in \mathbb{N}$ let $y_n \in f^{-1}(x_n)$ be given. It follows \begin{align*} (g \circ f)(y) = g(x_0) \neq \lim\limits_{n \to \infty} g(x_n) = \lim\limits_{n \to \infty} (g \circ f)(y_n). \end{align*} We have therefore shown that $g\circ f$ is discontinuous at $y$.

  • $\begingroup$ Who says $\lim_{n\to\infty}y_n$ has anything to do with $y$ (or $y_0$, you used both names for the same variable)? $\endgroup$
    – Ingix
    May 22, 2019 at 12:51
  • $\begingroup$ Without showing that $\lim_{n\to\infty} y_n=y_0$, this doesn't prove that $g\circ f$ is not continuous. $\endgroup$
    – 5xum
    May 22, 2019 at 13:07
  • $\begingroup$ I guess @neca has choosen the preimages wiseley and just forgot to tell us. $\endgroup$
    – Jupp
    May 22, 2019 at 15:15

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