Since the counterexample is about the distinction between $\cong$ being used for “the proposition that these two groups are isomorphic” and “the proposition that the implicitly specified map is an isomorphism”, I put additional language to clarify what is usually left implicit.

Anonymous

]]>Add counterexample that it is not enough to require each pair of homotopy groups be isomorphic.

Trebor

]]>I have added (here) the statement of Theorem 2 in Matumoto, Minami & Sugawara 1984, detecting weak homotopy equivalences on free homotopy sets.

They require their last condition for wedge sums of circles indexed by any set, which seems a weirdly strong condition. Inspection of the proof shows, unless I am missing something, that the actual index set being used is that underlying the fundamental group of the codomain space. So I have used this weaker condition in the proposition.

]]>added this reference:

- Takao Matumoto, Norihiko Minami, Masahiro Sugawara,
*On the set of free homotopy classes and Brown’s construction*, Hiroshima Math. J. 14(2): 359-369 (1984) (doi:10.32917/hmj/1206133043)

Added a remark about several other definitions of weak homotopy equivalences.

]]>would benefit from being split up into a few lemmas.

You could have a go at it in the $n$Lab entry. Would do the world a service, I am sure.

]]>I would definitely have an easier time understanding this lemma if it were stated using homotopy commutative diagrams, but I do think that its proof mixes up a few different lines of reasoning and would benefit from being split up into a few lemmas.

]]>The problem is that this “technical lemma” is awfully overloaded, in my mind this is five or six separate lemmas.

To me it is clearly a single lemma, but stated in a very confusing way because of the insistence on using one diagram that strictly commutes rather than talking about homotopies that live in squares and triangles.

]]>@14 They’re not quite dual. The notion of lifting property is self-dual, but here it splits into two distinct notions which are dual to each other: “lifting property up to under-homotopy” and “lifting property up to over-homotopy”. Morphisms characterized by such properties behave differently depending on whether they’re on the side of a strictly commuting triangle or on the side of a homotopy commuting one. The former are somewhat rigid and behave like (co)fibrations (for example Dold fibrations), the latter tend to be homotopy invariant and behave more like weak equivalences. I would be interested in learning about general theory of such lifting properties and classes of maps characterized by them, but I’m not aware of any such results.

@18 You’re right. The funny thing is that when I was first reading *A concise course in algebraic topology* a few years ago I spent a lot of time trying to understand what this lemma is about and finally gave up. Only now I realize that this is exactly the thing I learned to appreciate in the meantime. The problem is that this “technical lemma” is awfully overloaded, in my mind this is five or six separate lemmas.

Thanks for these pointers. I have added them here to the entry.

]]>but why complicate matters by bringing in the category of sets?

Only at the nForum would you hear a complaint that bringing in the category of sets complicates matters *vis a vis* the category of “$n$-tuply groupal sets”. (-:

@13 in the case of topological spaces is the “technical lemma” in section 9.6 of *A concise course in algebraic topology*. It, or something closely related to it, is also called the HELP lemma and apparently dates back at least to 1973, Boardman & Vogt “Homotopy invariant structures on topological spaces”. I agree that it is quite nice; it’s also quite similar to the solution-set condition in Smith’s theorem for generating combinatorial model category structures.

its being an isomorphism of such is equivalent to its being merely a bijection

Sure, but why complicate matters by bringing in the category of sets? If one is having trouble with the proposition that a certain group homomorphism is an isomorphism, then a handy theorem to use may be that it is so iff the underlying function is a bijection, but that fact is not the point.

It seems to me that the funny thing is that you can’t start any

lower

That’s because the concept of $(-1)$-tuply groupal set doesn’t make sense, as far as I know. (At least, it’s not in k-tuply monoidal n-category.) Thus, there is such a thing as $\Pi_{-1}(X)$ (which is a $(-1)$-groupoid), but no such thing as $\pi_{-1}(X,x)$ (which would be a $(-1)$-tuply groupal set).

]]>I have added a note under *Equivalent characterizations*.

Hi Karol,

that characterization is used quite widely at least in model category theory circles. I guess it was introduced in

- Jardine,
*Simplicial Presheaves*, Journal of Pure and Applied Algebra 47, 1987, no.1, 35-87.

Ah, that’s nice. And it is a vague sense dual to a trivial Dold fibration, where the homotopy is in the upper triangle, not the lower triangle.

]]>Maybe this is a good opportunity to advertise my favorite definition of a weak homotopy equivalence. A map of spaces $f : X \to Y$ is a weak homotopy equivalence if for every natural number $m$ and a diagram

$\begin{matrix} \partial I^m & \overset{u}{\longrightarrow} & X \\ \downarrow & & \, \downarrow f \\ I^m & \underset{v}{\longrightarrow} & Y \end{matrix}$there exist maps $w : I^m \to X$ and $H : I^m \times I \to Y$ such that $w | \partial I^m = u$ and $H$ is a homotopy from $f w$ to $v$ over $\partial I^m$. We obtain a definition of $k$-equivalence simply by restricting this condition to $m \le k$.

This definition is completely basepoint-free and it doesn’t refer to homotopy groups. In my experience it has the advantage that many basic properties of weak homotopy equivalences (and $k$-equivalences) can be verified using this definition without ever mentioning homotopy groups and the proofs tend to be neater than the ones that use the classical definition directly.

]]>Saying “isomorphism of n-tuply groupal sets” is not really necessary, though, since $\pi_n(f,x)$ is automatically a *morphism* of n-tuply groupal sets, so its being an isomorphism of such is equivalent to its being merely a bijection.

It seems to me that the funny thing is that you can’t start any *lower*:

- $\Pi_{-2}(f)\colon \Pi_{-2}(X) \to \Pi_{-2}(Y)$ is an equivalence of $(-2)$-groupoids, and $\pi_n(f,x)\colon \pi_n(X,x) \to \pi_n(Y,f(x))$ is an isomorphism of $n$-tuply groupal sets for all points $x$ of $X$ and all integers $n \geq -1$

is *not* a correct definition. Maybe this hierarchy should be discussed in the entry somewhere.

Well, sometimes people forget that for $n = 0$.

]]>If you don’t remember how π n(X,x) is an n-tuply groupal set for any natural number n,

If I didn’t remember that, I should go home and do something else than I am doing, such as Macrame maybe. ;-)

]]>Okay, so that first clause in your item (1) was missing!

]]>The following remark used to be in the entry, but I can’t see right now how it makes sense.

If it really doesn’t make sense to you, then I could explain; but if you just want to know why it’s in there at all, it’s because somebody (OK, it was me) *was* tempted to phrase the definition that way.

There’s actually a hierarchy of ways to phrase the definition:

- $\Pi_{-1}(f)\colon \Pi_{-1}(X) \to \Pi_{-1}(Y)$ is an equivalence of $(-1)$-groupoids, and $\pi_n(f,x)\colon \pi_n(X,x) \to \pi_n(Y,f(x))$ is an isomorphism of $n$-tuply groupal sets for all points $x$ of $X$ and all natural numbers $n \geq 0$;
- $\Pi_0(f)\colon \Pi_0(X) \to \Pi_0(Y)$ is an equivalence of $0$-groupoids, and $\pi_n(f,x)\colon \pi_n(X,x) \to \pi_n(Y,f(x))$ is an isomorphism of $n$-tuply groupal sets for all points $x$ of $X$ and all natural numbers $n \geq 1$;
- $\Pi_1(f)\colon \Pi_1(X) \to \Pi_1(Y)$ is an equivalence of $1$-groupoids, and $\pi_n(f,x)\colon \pi_n(X,x) \to \pi_n(Y,f(x))$ is an isomorphism of $n$-tuply groupal sets for all points $x$ of $X$ and all natural numbers $n \geq 2$;
- etc …;
- $\Pi_\infty(f)\colon \Pi_\infty(X) \to \Pi_\infty(Y)$ is an equivalence of $\infty$-groupoids, and [vacuous].

Starting too near the end begs the question when it comes to the homotopy hypothesis, while starting too near the beginning is unfamiliar. (I made the mistake of thinking that starting at the very beginning would simplify the phrasing, but it doesn’t.)

If you don’t remember how $\pi_n(X,x)$ is an $n$-tuply groupal set for any natural number $n$, this is covered at homotopy group (third Idea paragraph, and Examples in Low Dimensions).

]]>No, I erased my message (where I thought I had a simpler example than the pseudocircle) because it was in mathematical error! But since you ask, I think “reversible” gets the correct idea across.

]]>I didn’t see your original message, but here is a guess: Is maybe “reversible” too misleading, and too easily misread as “invertible”?

]]>(never mind)

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