Factorisation of continuous maps

Question

I'm studying general topology and a question has come to my mind.

I am referring to the class of theorems that in algebra go by the name of "homomorphism theorems". In my topology course, we have seen some results alike, but it lacks the general result on when, given two continuous maps with the same domain, there exists a continuous map which composed with the first gives the second. Is there a necessary and sufficient condition for that existence?

I know from set theory that for such a map to exist it's necessary that, if the first map has equal values on two arguments, so does the second. But there should be a second condition, involving the topological structure, assuring that the existing maps (possibly more than one, if the first map is not surjective) are continuous.

I add a statement highlighting the condition (?) which I'm looking for:

Let $f\colon X\to Y$ and $g\colon X\to Z$ be two continuous functions having the same domain (not only as sets, but as topological spaces). There exists a continuous function $h\colon Y\to Z$ such that $g = h\circ f$ if and only if (?).

Answer 1

You suggest a theorem of the form:

If $X$, $Y$, and $Z$ are structures, $f\colon X\to Y$ and $g\colon X\to Z$ are morphisms (respecting the structure), then provided that blah holds, there exists a morphism $h\colon Y\to Z$ such that $g=h\circ f$.

In groups, this is not a standard theorem (in the way that, say, the Noether Homomorphism Theorems are standard. Best I can come up with would be something like:

Proposition. If $f\colon G\to H$ and $g\colon G\to K$ are group homomorphisms, $\ker(f)\subseteq\ker(g)$, and $f$ is surjective, then there exists a group homomorphism $h\colon H\to K$ such that $g=h\circ f$.

(The condition $\ker(f)\subseteq\ker(g)$ is required because this is what ensures that $f(x)=f(x')$ implies $g(x)=g(x')$, which is required if $g=h\circ f$. Surjectivity of $f$ is also required, as for example the inclusion $f\colon\mathbb{Z}\hookrightarrow\mathbb{Q}$ and the identity map $g\colon\mathbb{Z}\to\mathbb{Z}$ cannot be factor into a map $\mathbb{Z}\hookrightarrow\mathbb{Q}\to\mathbb{Z}$.)

Far more common would be the First Isomorphism Theorem (or Fundamental Theorem of Homomorphisms):

Theorem. Let $f\colon G\to K$ be a group homomorphism. Then there exist a homomorphism $g\colon G/\ker(f)\to K$ such that $f=g\circ\pi$, where $\pi$ is the canonical projection; moreover, $g$ (co-)restricts to an isomorphism $G/\ker(f)\cong \mathrm{Im}(f)$.

There is such a theorem for topological spaces, using the notion of "quotient space".

Recall that if $(X,\tau)$ is a topological space, and $\sim$ is an equivalence relation on $X$, then the quotient space of $(X,\tau)$ by $\sim$ is the space with underlying set $X/\sim$ (the elements are the equivalence classes under $\sim$), and that has the final topology relative to the canonical projection $\pi\colon X\to X\sim$ (that is, the finest topology on $X/\sim$ that makes the canonical projection continuous).

Recall also that given any (set-theoretic) function $f\colon X\to Y$, the function $f$ induces an equivalence relation $\sim_f$ on $X$ by $x\sim_f y\iff f(x)=f(y)$.

Then we have:

Theorem. Let $X$ and $Y$ be topological spaces, and let $f\colon X\to Y$ be a continuous function. Then there exists a continuous function $g\colon X/\sim_f\to Y$ such that $f=g\circ\pi$, where $\pi\colon X\to X/\sim_f$ is the canonical projection.

Proof. We define $g\colon X/\sim_f\to Y$ by $g([x]) = f(x)$. This is well defined, since $[x]=[y]$ if and only if $f(x)=f(y)$. And trivially, $g\circ\pi=f$.

We show that $g$ is continuous: let $\mathscr{O}$ be an open set in $Y$. Then $g^{-1}(\mathscr{O})$ consists of all equivalence classes $[x]$ such that $f(x)\in\mathscr{O}$. This set is open in $X/\sim_f$ if and only if $\pi^{-1}(g^{-1}(\mathscr{O}))$ is open in $X$ (since the topology on $X/\sim_f$ is the final topology induced by $\pi$). And, indeed, we have $$\pi^{-1}(g^{-1}(\mathscr{O})) = (\pi^{-1}\circ g^{-1})(\mathscr{O}) = (g\circ\pi)^{-1}(\mathscr{O}) = f^{-1}(\mathscr{O}),$$ which is open in $X$ since $f$ is continuous. Thus, $g^{-1}(\mathscr{O})$ is open, and hence $g$ is continuous. $\Box$

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Suppose now that we have topological spaces $(X,\tau_X)$, $(Y,\tau_Y)$, and $(Z,\tau_Z)$, and $f\colon X\to Y$ and $g\colon X\to Z$ continuous functions with $f$ surjective. Under what conditions does there exist a continuous function $h\colon Y\to Z$ such that $g = h\circ f$?

We must have $\sim_f\subseteq\sim_g$. This is necessary and sufficient for the function $h$ to exist at the set-theoretic level. So it is necessary at the topological space level.

The resulting function $h$ must satisfy that $h(y) = g(x)$ where $x$ is any element of $X$ such that $f(x)=y$. So when is such an $h$ continuous?

Let $\mathscr{O}$ be an open set of $Z$. Then $y\in h^{-1}(\mathscr{O})$ if and only if $h(y)\in\mathscr{O}$. Then if $x\in X$ is such that $f(x)=y$, then $g(x)=h(y)\in\mathscr{O}$, so $x\in g^{-1}(\mathscr{O})$. Hence $y\in f(g^{-1}(\mathscr{O}))$. Thus, $h^{-1}(\mathscr{O})\subseteq f(g^{-1}(\mathscr{O}))$.

If $y\in f(g^{-1}(\mathscr{O}))$, then there exists $x\in g^{-1}(\mathscr{O})$ such that $f(x)=y$. So $h(y) = h(f(x)) = g(x)\in\mathscr{O}$. So $f(g^{-1}(\mathscr{O}))\subseteq h^{-1}(\mathscr{O})$. Thus, $h^{-1}(\mathscr{O}) = f(g^{-1}(\mathscr{O}))$.

Let $\tau_g$ be the initial topology on $X$ induced by $g$; that is, $\tau_g$ is the coarsest topology for which $g$ is continuous; in particular, we must have that $\tau_X\subseteq\tau_g$. Note that $\tau_g$ consists precisely of the sets $g^{-1}(\mathscr{O})$ where $\mathscr{O}\in\tau_Z$ is an open subset of $Z$. So in order for $h$ to be continuous, we require that $f\colon (X,\tau_g)\to Y$ be an open mapping. So:

Theorem. Let $(X,\tau_X)$, $(Y,\tau_Y)$, $(Z,\tau_Z)$ be topological spaces, let $f\colon X\to Y$ be a surjective continuous function and let $g\colon X\to Z$ be a continuous. Let $\tau_g$ be the coarsest topology on $X$ that will make $g$ continuous, and let $\sim_f$ and $\sim_g$ be the equivalence relations on $X$ induced by the functions $f$ and $g$. Then there exists a continuous function $h\colon Y\to Z$ such that $g=h\circ f$ if and only if the following two conditions hold:

1. $\sim_f\subseteq \sim_g$; that is, $f(x)=f(x')\implies g(x)=g(x')$; and
2. $f(\tau_g)\subseteq \tau_Y$; that is, the direct image of every open set in $\tau_g$ is open in $Y$.

Proof. Assume $h$ exists. Then $f(x)=f(x')\implies g(x)=h(f(x))=h(f(x'))=g(y)$, so $\sim_f\subseteq\sim_g$. This proves 1.

Let $U\in\tau_g$. Then there exists an open set $\mathscr{O}\in \tau_Z$ such that $U=g^{-1}(\mathscr{O})$. Then, as above, we have that $h^{-1}(\mathscr{O})=f(U)$. Since we are assuming that $h$ is continuous, it follows that $h^{-1}(\mathscr{O})\in \tau_Y$, hence $f(U)\in\tau_Y$. This proves 2.

Conversely, assume that 1 and 2 hold. We define $h\colon Y\to Z$ as follows: if $y\in Y$, then there exists $x\in X$ such that $f(x)=y$. Define $h(y)=g(x)$. Then 1 yields that $h$ is well defined, and hence $g=h\circ f$ To show $h$ is continuous, let $\mathscr{O}$ be open in $Z$. Then as above we have $h^{-1}(\mathscr{O})=f(g^{-1}(\mathscr{O}))$. Since $g$ is continuous, then $g^{-1}(\mathscr{O})\in \tau_X\subseteq\tau_g$, hence $f(g^{-1}(\mathscr{O})\in\tau_Y$ by 2; thus, $h^{-1}(\mathscr{O})$ is open in $Y$, proving that $h$ is continuous. $\Box$

I am not sure what a condition would look like if we drop the requirement that $f$ be surjective. It is clear we can construct $h$ provided that the conditions above hold and in addition $f(X)$ is clopen (just map all of $Y\setminus f(X)$ to a single point). But if it is not clopen (it must be open in order to satisfy 2) then I'm not sure.

Answer 2

Let $X$ be a compact Hausdorff space, [EDIT: $Y$ and $Z$ Hausdorff], $f: X \to Y$ and $g: X \to Z$ continuous such that $f(x_1) = f(x_2)$ implies $g(x_1) = g(x_2)$. This condition lets us define $h: f(X) \to Z$ by $h(y) = g(x)$ where $f(x) = y$. Thus $h \circ f = g$. The question is whether $h$ is continuous.

If not, there is a net $y_\alpha$ in $f(X)$ converging to some $y\in f(X)$ with $h(y_\alpha)$ not converging to $h(y)$. Taking a subnet, by compactness of $X$ we may assume $y_\alpha = f(x_\alpha)$ with $x_\alpha$ converging to some $x \in X$. By continuity of $f$, $y = \lim_\alpha y_\alpha = \lim_\alpha f(x_\alpha) = f(x)$. So $h(y) = g(x)$. But by continuity of $g$, $g(x) = \lim_\alpha g(x_\alpha) = \lim_\alpha h(y_\alpha)$.