# Which maps of topological spaces have the right lifting property with respect to all split monomorphisms?

Let $$p : X \to Y$$ be a continuous map. We say that $$p$$ has the right lifting property with respect to split monomorphisms if, for every space $$B$$, and every retract $$A \subseteq B$$, and for every continuous $$u : A \to X$$, $$v : B \to Y$$ such that $$p\circ u = v|_A$$, there exists a continuous $$w : B \to X$$ such that $$w|_A = u$$ and $$p\circ w = v$$.

Question: Which maps $$p : X \to Y$$ have the right lifting property with respect to split monomorphisms?

Note that when testing the right lifting property, we may assume without loss of generality that $$A = X$$ and $$u : A \to X$$ is the identity. So the property equivalently says that if $$X$$ is a retract of $$B$$ and there is a map $$v : B \to Y$$ such that $$v|_X = p$$, then there exists a choice of retraction $$w : B \to X$$ such that $$pw = v$$.

Notes:

• I'm happy to restrict to some convenient category of topological spaces.

• The split monomorphisms are closed under cobase-change, (transfinite) composition, and retracts, so modulo size issues (which I think are potentially serious) there should be a weak factorization system on $$Top$$ with left class the split monomorphisms.

• Moreover, the cograph factorization gives us (split mono, split epi factorizations). This is the factorization of a map $$f : X \to Y$$ as $$X \to X \amalg Y \to Y$$.

• Using the cograph factorization, we see that if $$p : X \to Y$$ has the right lifting property with respect to split monomorphisms, and if $$X \neq \emptyset$$, then $$p$$ is a split epimorphism, i.e. there is a map $$s : Y \to X$$ with $$p \circ s = id_X$$.

• But I suspect that not every split epimorphism has the right lifting property with respect to split monomorphisms.

• I'm also curious about the dual question: which maps $$i : A \to B$$ have the left lifting property with respect to split epimorphisms? I'd also be curious about any general remarks which can be made about this situation in other categories. None of the observations I've made above use anything special about $$Top$$.

• Why is the "cograph" not isomorphic to $Y$? (In the same way that the graph of a map is isomorphic to its domain). Commented Aug 31, 2023 at 21:42
• @MaximeRamzi Because every relation that defines a function has injective projection from $X \times Y$ to $X$, but not necessarily surjective projection to $Y$. Commented Aug 31, 2023 at 22:05
• @MaximeRamzi Because I messed it up -- fixed now! The cograph, as an object, is just $X \amalg Y$. Commented Aug 31, 2023 at 22:22
• Minor remark: If $X$ is empty and $Y$ is not, then $X\to X\amalg Y$ is not split mono. But of course in that case you still have a (split mono,split epi) factorization $X\to X\to Y$. Commented Sep 1, 2023 at 10:53
• If $p$ has the RLP w.r.t. all split monomorphisms, then it must be a Hurewicz fibration, since $A\times I$ has $A\times 0$ as a retract. Commented Sep 1, 2023 at 11:48

SEE THE EDITS BELOW

The answer is basically: fibrations equivalent to trivial bundles. (Dylan Wilson speculated about this in the [comments (https://mathoverflow.net/questions/453780/which-maps-of-topological-spaces-have-the-right-lifting-property-with-respect-to#comment1174152_453780).) But I am not sure what exactly I should mean by "basically", or "equivalent".

As I pointed out in a comment, $$p$$ must be a (Hurewicz) fibration because $$A\times 0$$ is a retract of $$A\times I$$ for every space $$A$$.

Suppose that $$p:X\to Y$$ has the RLP w.r.t. all split monomorphisms, and assume that $$Y$$ is path-connected. Let $$y\in Y$$ be a point. Let $$F\subset X$$ be the fiber over $$y$$.

The injection $$i:F\to F\times Y$$ given by $$f\mapsto (f,y)$$ is a split monomorphism. Define $$u:F\to X$$ by inclusion, and define $$v:F\times Y\to Y$$ by projection. Note that both $$p\circ u$$ and $$v\circ i$$ are the constant map at $$y$$.

A solution $$w:F\times Y\to X$$ of the lifting problem is a map, over $$Y$$, between the trivial bundle $$F\times Y\to Y$$ and the fibration $$p:X\to Y$$, such that on the fiber over $$y$$ it is the identity map of $$F$$. Therefore over every point in $$Y$$ the map of fibers is a homotopy equivalence, and therefore the map $$F\times Y\to X$$ is a homotopy equivalence.

EDITED: Here is a definite characterization of maps having the RLP w.r.t. all split monomorphisms: they are the retracts of trivial fiber bundles. That is, $$p:X\to Y$$ has this property if and only if there exists some space $$Z$$ such that, in the category of all spaces over $$Y$$, $$p$$ is a retract of the product projection $$Z\times Y\to Y$$.

Proof of "if" direction: Any right lifting property is inherited by retracts and also by base change, so it is enough that $$Z\times \ast$$ has the RLP w.r.t. split monomorphisms. And in fact split monomorphisms are precisely the maps that have the LLP w.r.t. $$Z\to \ast$$ for every $$Z$$.

Proof of "only if" direction: Let $$Z$$ be $$X$$. The map $$i:X\to X\times Y$$ given by $$i(x)=(x,p(x))$$ has left inverse given by $$(x,y)\mapsto x$$, so it is split mono. We show that $$X$$, as a space over $$Y$$, is a retract of $$X\times Y$$, by producing a different left inverse for $$i$$, a map $$w: X\times Y\to X$$ that preserves the map to $$Y$$. For this we use that $$p:X\to Y$$ has the RLP w.r.t. $$i:X\to X\times Y$$:

Let $$u:X\to X$$ be the identity, and let $$v:X\times Y\to Y$$ be projection. Then $$v\circ i=p=p\circ u$$, so there exists $$w$$ such that $$w\circ i=u$$ and $$p\circ w=v$$.

QED

It seems funny that (as discussed in the first part of the answer) a retract of a trivial bundle should always be at least weakly fiberwise homotopy equivalent to another trivial bundle, as long as the base is path-connected. But it's true.

Specifying a retract of $$Z\times Y$$ (as a space over $$Y$$) means given a map $$e:Z\times Y\to Z\times Y$$ such that $$e\circ e=e$$ and such that the projection to $$Y$$ is preserved, in other words giving a family $$e_y$$ of idempotent self-maps of $$Z$$, parametrized by $$Y$$. (The fibers over $$Y$$ of) the image $$E=e(Z\times Y)$$ may be thought of as a family $$F_y=e_y(Z)$$ of retracts of $$Z$$ parametrized by $$Y$$. Fixing some $$y_0\in Y$$, we can map $$F_{y_0}\times Y$$ into $$E$$ by $$e$$, and this map of fibrations over $$Y$$ is a homeomorphism in the fiber over $$y_0$$ and so must be a weak equivalence in every fiber.

I informally think of this conclusion in the following way: The canonical map from the space of all idempotent self-maps of $$Z$$ to the space of all retracts of $$Z$$ is such that on each component it is homotopic to a constant map.

By the way, if $$Y$$ is discrete then a map $$X\to Y$$ has this RLP if and only if either all the fibers are empty or all the fibers are nonempty.

EDIT About the dual question: A map $$A\to B$$ of spaces has the LLP with respect to all split epimorphisms if and only if it is an open and closed embedding, i.e. iff it corresponds to a homeomorphism from $$A$$ to an open and closed subset of $$B$$. You can see this by using the evident split epimorphism $$A\amalg B\to B$$.

• Although it seems to me that the class of (regular) fibre-homotopically trivial Hurewicz fibrations is larger than the class of maps which Tim is interested in. Commented Sep 2, 2023 at 14:01
• @Tyrone: Yes, I think I agree. Commented Sep 2, 2023 at 14:28
• Yes, I was loosely using "trivial fibration" to mean "fibration that is in some sense equivalent to a trivial bundle". Commented Sep 2, 2023 at 20:53
• Thanks, this is really nice! Using the graph factorization, we see that (split monos, retracts of product projections) is a weak factorization system on $Top$, or in fact on any category with binary products. And in $Top$, it is very good to know that a retract of a product projection is fiberwise homotopy equivalent to a product projection. Commented Sep 3, 2023 at 14:45
• Regarding the dual situation, I suppose another way of saying this is that the coproduct inclusions in $Top$ are closed under retracts. Commented Sep 5, 2023 at 17:32