The continuity of $g\circ h$ and $h$, and $h$ surjective, imply the continuity of $g$?

Question

Structural assumptions: $X$ is a nontrivial connected separable topological space

Given: $f(x)=g(h(x))$.

$f:X\to\mathbb R$ is continuous

$h:X\to\mathbb R$ is continuous and $h(X)=\mathbb R$.

$g:\mathbb R\to\mathbb R$.

Question: Can we say $g$ is also continuous?

Motivation: Composition of continuous functions is also continuous. Does an "inverse" argument holds?

After doing some research the argument seems to be correct for metric spaces with additional assumptions of one of the function being homeomorphism: If the composition of two functions is continuous and one of those functions is continuous, is the other function continuous as well?

Note: superscripts in this proof are indexes not exponents.

My test: I think if we let $X$ be path-connected, it is easy to prove that $g$ is continuous. Let $y_1<y_2$ be two real numbers and $h(x_i)=y_i$, for any real number $y\in (y_1,y_2)$, consider a sequence $y^n\in (y_1,y_2)$ converging to $y$. We will show that $g$ is continuous at arbitrary real number $y$.

Let $T$ be the path between $x_1,x_2$. It follows from the continuity of $h$ that for each $n$ there exists $x^n\in T$ such that $h(x^n)=y$. Because of path-connectedness, path $T$ is compact. So sequence $x^n$ converges to $x$ where $f(x)=y$. By the continuity of $f$, it follows that $f(x^n)$ converges to $f(x)$ and $g(h(x^n))$ converges to $g(h(x))$. Finally, we conclude that $h(x^n)\to h(x)$ $\implies g(h(x^n))\to g(h(x))$.

However, I think this method only works for path-connected $X$ not connected $X$. Could you please give me some hints or a counterexample?

Answer 1

[Updated with significantly simplified counterexample.]

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No, we cannot say $g$ is continuous for the general connected case. We have a counterexample already with $X\subseteq \mathbb R^2$.

Let $Y^0\subseteq \mathbb R^2$ be the $y$-axis with the origin removed, and for each $n\in\mathbb Z\backslash \{0\}$ let $J_n=(0,1)\times\{\frac{1}{n}\}$. Denote also $J_\infty= (0,1)\times\{0\}$.

Let $$X=Y^0\cup J_\infty\cup \bigcup_{n\neq 0} J_n,$$ and note that $X$ is connected but not path connected.

Let $h\colon X\to\mathbb R$ be the $y$-coordinate function restricted to $X$, (which is clearly continuous and surjective) and define $g\colon \mathbb R\to \mathbb R$ by $$g(t)= \begin{cases} \sin(\frac{\pi}{t}) &t\neq 0\\ 0 & t=0 \end{cases} $$

so that $g$ is discontinuous at $0$ but continuous everywhere else.

Now $f=g\circ h$ is certainly continuous on each $J_n$, including $J_\infty$. To see this, note that the composition is identically equal to $0$ on the union of these sets, which is open in $X$ (in fact it is given by $X\cap\{(x,y)\in\mathbb R^2\mid x>0\}$). Moreover, $f$ is continuous at each point in $Y^0$, since $h$ is nonzero there, and $g$ is continuous away from $0$. Hence $f$ is continuous at every point.

Answer 2

This is not an answer to the question, but a remark which is too long for a comment. Let me give a correct proof of the statement for path-connected $X$.

Pick any point $y_0\in \mathbb{R}$. We want to prove that $g$ is continuous at this point. For this we pick a sequence $(y_n)_{n\geq 1}\subseteq \mathbb{R}$ such that $y_n \rightarrow y_0$ as $n\rightarrow \infty$. We would like to pick a point $x_0\in X$ such that $g(x_0)=y_0$ and for each $n$, $x_n\in h^{-1}(\{y_n\})$ such that $x_n \rightarrow x_0$. In that case we would get $$ g(y_n)=h(g(x_n))\rightarrow h(g(x_0))=h(y_0),$$ where we used that continuous functions are sequentially continuous (see here https://math.stackexchange.com/questions/2343261/sequentially-continuous-implies-continuous#:~:text=Continuity%20always%20implies%20sequential%20continuity,continuity%20of%20f) to conclude that $g$ is continuous at $y_0$ and hence (as $y_0$ was arbitrary) that $g$ is continuous.

Our job is now to find $x_0$ and $(x_n)_{n\geq 1}$ as above. As $h$ is surjective, there exist $z_0, z_1\in X$ such that $h(z_0)=y_0+1$ and $h(z_1)=y_0-1$. Furthermore, as $X$ is path-connected, there exists a continuous path $\gamma:[0,1]\rightarrow X$ such that $\gamma(0)=z_0, \gamma(1)=z_1$. Next we note that $\varphi:=h\circ \gamma: [0,1]\rightarrow \mathbb{R}$ is continuous and thus, by the intermediate value theorem $\varphi([0,1])\supseteq [y_0-1,y_0+1]$. As $y_n \rightarrow y_0$ we can wlog assume that $(y_n)_n \subseteq \varphi([0,1])$. Now pick $x_n \in \gamma([0,1])$ such that $h(x_n)=y_n$. Using the sequential compactness of $\gamma([0,1])$ we get that for every subsequence $(x_{n_m})_m$ there exists a subsequence $(x_{n_{m_\ell}})_\ell$ and some point $x_0\in \gamma([0,1])$ such that $x_{n_{m_\ell}} \rightarrow x_0$. We have $$h(x_0)=\lim_{\ell\rightarrow \infty} h(x_{n_{m_\ell}})=\lim_{\ell\rightarrow \infty}y_{n_{m_\ell}} = y_0.$$ Thus, we get $$ \lim_{\ell \rightarrow \infty} g(y_{n_{m_\ell}}) = \lim_{\ell \rightarrow \infty} g(h(x_{n_{m_\ell}})) = g(y_0). $$ We have shown that for every subsequence $(g(y_{n_m}))_m$ there exists a subsequence $(g(y_{n_{m_\ell}}))_\ell$ such that $$ \lim_{\ell \rightarrow \infty} g(y_{n_{m_\ell}}) = g(y_0). $$ Therefore, Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.,
$$ \lim_{n\rightarrow \infty} g(y_n)=g(y_0) $$ and $g$ is continuous at $y_0$ (here I assumed that $\mathbb{R}$ comes with the Euclidean topology in order to get that sequentially continuous maps are continuous, this would work with any first-countable topology).

Added: We could also drop the assumption that $X$ is path-connected and assume instead that $g$ is a proper map (that means the preimages of compact sets under $g$ are again compact) and that $X$ is sufficiently nice to get the equivalence of compactness and sequential compactness. The proof above goes through as soon as we know that $h^{-1}([y_0-\varepsilon, y_0+\varepsilon])$ is sequentially compact for some $\varepsilon>0$.

Answer 3

I'm a little confused at which hypotheses are being placed on the spaces involved, but if $X, Y$ and $Z$ are path-connected metric spaces, and we are given functions $h\colon X \to Y$ and $g\colon Y \to Z$ such that $h$ is continuous and surjective, and the composition $g\circ h$ is continuous, then it does not follow that $g$ is continuous:

Example: Take $X=[0,1)$ and let $h\colon X\to S^1 = \{z \in \mathbb C: |z|=1\}$ be given by $h(t) = \exp(2\pi i t)$. Then $X$ and $S^1$ are Hausdorff and path-connected and $h$ is a continuous surjection. But if we let $g\colon S^1 \to X$ where $g(z) = t\in[0,1)$ and $h(t)=\exp(2\pi i t)=z$, i.e. $g$ is the set-theoretic inverse of $h$, then we have $g\circ h(t)=t$ for all $t \in X$ and hence $g\circ h$ is continuous, but clearly $g$ fails to be continuous at $z=1 \in S^1$.

Update: The above example generalises to give a necessary and sufficient condition for when a map $h\colon X\to Y$ has the property that for any $g\colon Y \to Z$, $g$ is continuous if and only if $g\circ h$ is continuous precisely when $h$ is a quotient map.

Definition: If $\sim$ is an equivalence relation on a topological space $X$, then we let $X/\sim$ be the set of equivalence classes of $\sim$ and $q \colon X \to X/\sim$ the quotient map sending $x\in X$ to its equivalence class $[x]$. The set of equivalence classes $X/\sim$ has a natural topology, namely we say $U\subseteq X/\sim$ is open precisely when $q^{-1}(U)$ is open. This is the finest topology on $X/\sim$ for which the map $q$ is continuous. If $f\colon X \to Y$ is any function which satisfies $f(x)=f(x')$ whenever $x\sim x'$, then $f$ induces a unique function $\tilde{f}\colon X/\sim \to Y$ satisfying $\tilde{f}\circ q = f$, and $f$ is continuous if and only if $\tilde{f}$ is continuous.

Given any continuous function $f\colon A \to B$ of topological spaces there is an associated equivalence relation $\sim$ on $A$ given by $a_1 \sim a_2$ if $f(a_1)=f(a_2)$. Clearly if $\tilde{A}= A/\sim$ then $f$ factors through $q\colon A \to \tilde{A}$, say $f = \tilde{f}\circ q$. We say that $f$ is a quotient map if $\tilde{f}\colon \tilde{A} \to B$ is a homeomorphism.

Proposition: If $h\colon X \to Y$ is a continuous map then, for any $g\colon Y \to Z$ we have $g\circ h$ continuous if and only if $g$ is continuous precisely when the map $h$ is a quotient map.

Proof: We must show that $$ g\circ h \text{ is continuous} \implies g \hspace{1mm} \text{is continuous} $$ holds if and only if $h$ is a quotient map. Let $q\colon X \to \tilde{X}=X/\sim$ be the quotient map where $\sim$ is the equivalence relation $x\sim x'$ if $h(x)=h(x')$. Now any function $f \colon X \to Z$ which satisfies $f(x)=f(x')$ whenever $x\sim x'$ induces a unique function $\tilde{f}\colon \tilde{X} \to Z$ satisfying $f= \tilde{f}\circ q$, and $f$ is continuous if and only if $\tilde{f}$ is. Thus in particular $h=\tilde{h}\circ q$ where $\tilde{h}\colon \tilde{X} \to Y$ is a continuous bijection, and if $g\colon Y \to Z$ is a function, $g\circ h = g\circ \tilde{h}\circ q$ hence $\widetilde{g\circ h} = g\circ \tilde{h}$, and so $g\circ h$ is continuous if and only if $g\circ \tilde{h}$ is.

But now if $h$ is a quotient map, then by definition $\tilde{h}$ is a homoemorphism, i.e. $\tilde{h}^{-1}$ is continuous. But then if $g\circ \tilde{h}$ is continuous, $g = (g\circ \tilde{h})\circ \tilde{h}^{-1}$ is continuous, so the condition that $h$ is a quotient map is certainly sufficient. Conversely if $h$ is not a quotient map, then $g=\tilde{h}^{-1}\colon Y \to \tilde{X}$ is not continuous, while $h \circ \tilde{h}^{-1} = q$ is continuous, so that the continuity of $h\circ g$ does not imply the continuity of $g$ for all $g\colon Y \to Z$.

Lemma Let $h\colon X \to Y$ be a continuous surjection onto a Hausdorff space $Y$. If there exist compact subsets $\{C_i:i \in I\}$ of $X$ such that $\{\mathrm{int}(h(C_i)): i \in I\}$ form a covering of $Y$ by open sets, then $h$ is a quotient map.

Proof: We must show that, given such a collection $\{C_i: i \in I\}$ of compact subsets of $X$, if $F \subseteq Y$ and $h^{-1}(F)$ is closed, then $F$ is closed in $Y$. To see this, note that if $h^{-1}(F)$ is closed in $X$ then for each $i \in I$ the intersection $C_i \cap h^{-1}(F)$ is closed in $C_i$, and hence is compact. Thus $h(C_i\cap h^{-1}(F)))=h(C_i) \cap F$ is compact and therefore closed (as $Y$ is Hausdorff) in $Y$. But then $$ F\cap \mathrm{int}(h(C_i)) = h(C_i\cap h^{-1}(F))\cap \mathrm{int}(h(C_i)) $$ is closed in $\mathrm{int}(h(C_i))$. Since $\bigcup_{i\in I} \mathrm{int}(h(C_i))=Y$, it follows that $F$ is closed in $Y$ as required. (Check this!)

Remarks:

1. In particular, if $X$ is compact, then any continuous surjection $h\colon X \to Y$ to a Hausdorff space $Y$ is always a quotient map.

2. The conditions that $X$ is path-connected and $Y=\mathbb R$ give a less trivial setting where the hypothesis of the previous Lemma holds: If $y_0 \in \mathbb R$, we may find $z_0,z_1 \in X$ with $h(z_0)= y_0-1$, and $h(z_1) = y_0+1$. Then since $X$ is path-connected, there is a path $\gamma\colon [0,1]\to X$ with $\gamma(0)=z_0$ and $\gamma(1)=z_1$. The composition $\phi=h \circ \gamma \colon [0,1]\to \mathbb R$ is continuous, and hence the Intermediate Value Theorem ensures that $(y_0-1,y_0+1)\subseteq \mathrm{int}(h(\gamma([0,1]))$, and $\gamma([0,1])=C_{y_0}$ is a compact subset of $X$, so that $\{C_{y_0}: y_0 \in Y\}$ is a collection of compact subsets of $X$ satisfying the hypothesis of the Lemma as required.

3. The Proposition shows that the connectedness of $X$ is not really related to the question of whether $h\colon X \to Y$ can determine whether or not a function $g\colon Y \to Z$ is continuous: Given a quotient map $h\colon X \to Y$ with $X$ connected, if $\mathcal U = \{U_i: i \in I\}$ is an open cover of $X$, then setting $X_{\mathcal U} = \bigsqcup U_i\times I \subseteq X\times I$ (where we give $I$ the discrete topology) and $\pi\colon X_{\mathcal U} \to X$ the restriction of the first projection $X\times I \to X$ to $X_{\mathcal U}$ it is easy to see that $\pi$ is a quotient map and so $h\circ \pi$ is a quotient map, but $X_{\mathcal U}$ is only connected if $\mathcal U = \{X\}$, thus the failure of $X$ to be connected in no way precludes a map $h\colon X \to Y$ from determining the continuity of maps from $Y$.

4. The example given by M.W. of a continuous map $h\colon X \to Y$ from a connected $X$ to $Y=\mathbb R$ which fails to determine the continuity of maps $g\colon Y \to Z$ is not a quotient map, because, for example, if $F= \{\frac{1}{n\pi}: n \in \mathbb N\}$ then $F$ is clearly not closed in $Y=\mathbb R$, but $h^{-1}(F)= \{(0,\frac{1}{n\pi}): n \in \mathbb N\}$ is closed in $X$.