$\let\op=\operatorname$In $\op{Set}$, we have an $(\op{Epi},\op{Mono})$ orthogonal factorization system. Strikingly, if we reverse the roles, we get the no-less-important $(\op{Mono},\op{Epi})$ weak factorization system.

In the $\infty$-category $\op{Spaces}$ of spaces, the most direct analog of the $(\op{Epi},\op{Mono})$ orthogonal factorization system on $\op{Set}$ is the $(\text{Effective Epi}, \op{Mono})$ orthogonal factorization system, but this is just the $(-1)$th in a whole tower: for each $n \in \mathbb Z_{\geq -2}$, we have an $(\text{$n$-connected}, \text{$n$-truncated})$ factorization system [1].

It seems that, just as in the analogous case in $\op{Set}$, one can take the left half of each these orthogonal factorization systems, and view it as the right half of a weak factorization system $(\mathcal L_n, \text{$n$-connected})$ [2]. To see this, one shows that the $n$-connected maps are precisely the maps which are weakly right orthogonal to the maps $\{S^k \to 1 \mid -1 \leq k \leq n\}$, and applies the small object argument to obtain factorizations.

In $\op{Set}$, we have the cute fact that the resulting weak factorization system $(\op{Mono},\op{Epi})$ is just the original orthogonal factorization system $(\op{Epi},\op{Mono})$ with the left and right classes swapped. This is not the case in $\op{Spaces}$, even when $n=-1$: a map $A \xrightarrow i B$ of spaces is weakly left orthogonal to the effective epimorphisms if and only if it is a coproduct inclusion $A \to A \amalg S$ where $S$ is discrete; this is more restrictive than being a monomorphism [3]. I don't know how to characterize the left class $\mathcal L_n$ for $n\geq 0$ as cleanly. In fact, unlike the case in $\op{Set}$, I don't think we have either containment $\mathcal L_n \subseteq \text{$n$-truncated}$ or $\text{$n$-truncated} \subseteq \mathcal L_n$ in general. This leads to my

Questions: Let $n \in \mathbb Z_{\geq -2}$.

  1. Is there a good characterization of the class of maps $\mathcal L_n$, i.e., the maps of spaces which are weakly left orthogonal to the $n$-connected maps?

  2. What would be a good name for the maps of $\mathcal L_n$?

Note that by the small object argument, the maps of $\mathcal L_n$ are precisely the retracts of transfinite composites of cobase-changes of coproducts of the maps $\{S^k \to 1 \mid -1 \leq k \leq n\}$. So in some sense this is a quite explicit class of maps. By "characterization" I suppose I mean something which can be "checked directly" without having to find all the data of a construction of this form.

[1] Here a map is said to be $n$-truncated or $n$-connected if its fibers are all so. This convention is off by one from the most classical convention.

[2] Some care should be taken with the definition of a weak factorization system $\infty$-categorically: say that a morphism $A \xrightarrow i B$ is weakly orthogonal to a morphism $X \xrightarrow p Y$ if the map $\op{Hom}(B,X) \to \op{Hom}(B,Y) \times_{\op{Hom}(A,Y)} \op{Hom}(A,X)$ is an effective epimorphism. Spelled out, this says that if we have a commutative square—i.e., morphisms $A \xrightarrow u X$, $B \xrightarrow v Y$ along with a homotopy $\gamma: pu \sim vi$, then there exists a lift, i.e., $B \xrightarrow w X$ and homotopies $\alpha: wi \sim u$, $\beta: pw \sim v$ and (here's the only subtle part) a homotopy of homotopies from the composite $\beta \ast \alpha$ to $\gamma$. Then a weak factorization system is, as usual, a pair of classes of morphisms $(\mathcal L, \mathcal R)$ which are complements to each other with respect to weak orthogonality, such that every morphism admits a factorization as a morphism in $\mathcal L$ followed by a morphism in $\mathcal R$.

[3] Recall that a monomorphism of spaces is a coproduct inclusion $A \to A \amalg S$ where $S$ may be an arbitrary space.

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    $\begingroup$ Good grief. $\mathcal L_n$ is just the class of maps which are retracts of relative cell complexes of dimension $\leq n+1$. I'd probably call these "retracts of $n+1$-skeletal maps" or something. $\endgroup$ – Tim Campion Feb 15 at 18:25
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    $\begingroup$ Is this your answer to your own question? I have trouble parsing the meaning of "good grief" here …. $\endgroup$ – LSpice Feb 15 at 22:17
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    $\begingroup$ @LSpice. Yeah, I think I answered my own question... Maybe I should delete it, but I spent so much time typing it out... And now you've invested time in prettying it up, too! (Thanks, btw) Actually, I suppose it's still an interesting question how to characterize these maps, but $\mathcal L_n$ is so familiar that if a good characterization were known it would have been known in the '50's, and I don't think such a thing is known... Although maybe in the simply-connected case you can say something about dimension vs. homology dimension or something... $\endgroup$ – Tim Campion Feb 15 at 22:18
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    $\begingroup$ To clarify, "Good grief" is my sheepish exclamation when I realized that I found a ridiculous roundabout way to ask for a characterization of the retracts of relative $n+1$-dimensional complexes without recognizing that that's what I was asking! $\endgroup$ – Tim Campion Feb 15 at 22:28
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    $\begingroup$ Yes, there is a characterization in terms of (co)homology in most cases. Of course for a map in this class the relative homology and cohomology vanishes in degrees above $n+1$. Conversely, if relative $H^j$ vanishes for all $j\ge n+1$, for all coefficient systems on the codomain, then the map is in that class. This holds for all $n>2$, at least. $\endgroup$ – Tom Goodwillie Feb 15 at 23:46

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