3 added 27 characters in body

I was thinking of asking a similar question. This won't be an answer, but it might be a slightly useful reframing. Presumably an expert will give us a real answer.

For spectra $X$ and $Y$, let's write $X\perp Y$ if the mapping spectrum $Hom(X,Y)$ is contractible, i.e. if the cohomology $Y^\star X$ is trivial, i.e. if in the stable homotopy category there is no nontrivial map $X\to \Sigma^nY$ for any $n\in\mathbb Z$.

Consider pairs $(\mathcal C,\mathcal L)$ of classes of stable homotopy types such that ($X\in \mathcal C$ iff $X\perp Y$ for all $Y\in \mathcal L$) and ($Y\in \mathcal L$ iff $X\perp Y$ for all $X\in \mathcal C$). Let's call such a thing a matched pair, just to have a term for it.

The best matched pairs are those which have localization functors. This means that for every spectrum $X$ there is a universal example of a map $X\to L$ from $X$ to an element of $\mathcal L$.

Here are some properties that a class $C$ of stable homotopy types may or may not have:

1. $\mathcal C$ is the first member of a localization pair $(\mathcal C,\mathcal L)$.

2. $\mathcal C$ is the first member of a matched pair $(\mathcal C,\mathcal L)$.

3. There is a class $\mathcal E$ of spectra such that $X\in \mathcal C$ iff for every $E\in \mathcal E$ we have $E_\star X=0$.

4. $\mathcal C$ is closed under homotopy colimits and desuspension.

Of course 1 implies 2. The converse seems like a good question.

The question of whether a given matched pair $(\mathcal C,\mathcal L)$ is a localization pair can be reformulated in a couple of ways. Given a fibration sequence $\star\to C\to X\to L\to \star$, the following three conditions are equivalent:

(a) $L\in \mathcal L$ and $X\to L$ is universal among maps from $X$ to objects in $\mathcal L$.

(b) $C\in \mathcal C$ and $L\in \mathcal L$.

(c) $C\in \mathcal C$ and $C\to X$ is universal among maps to $X$ from objects in $\mathcal C$.

Clearly 2 implies 4 and 3 implies 4.

3 is like saying that $\mathcal C$ is a Bousfield acyclicity class, except that we are allowing $\mathcal E$ to be a proper class instead of a set.

(In the same sense 2 says that $\mathcal C$ is what might be called a cohomological Bousfield class.)

If we take a "What, me worry?" attitude toward set theory, then 3 implies 1: every Bousfield class has a localization functor.

In fact it seems that within the framework of this "Alfred E. von Neumann set theory" 4 implies 1, by the usual argument (carried out rigorously by Bousfield with appropriate hypotheses): To create a localization of $X$, start throwing elements of $\mathcal C$ at it and take cofibers, and keep doing this transfinitely until you are done. The resulting spectrum $L$ should belong the class $\mathcal L$ of all spectra that satisfy $C\perp L$ for all $C\in\mathcal C$. The (co)fiber of the map $X\to L$ is made from $\mathcal C$ by repeated colimits, and so is in $\mathcal C$. This makes $X\to L$ universal. Applying that same construction when $X$ satisfies $X\perp L$ for $L\in \mathcal L$, we find that such $X$ is a retract of an object in $\mathcal C$, therefore is in $\mathcal C$.

So modulo set theory 1=2=4.

But as far as I know the question of whether 3 is strictly stronger is of a different nature.

About (co)reflexivity: 1 implies that $\mathcal C$ is coreflexive. To get a converse you have to assume more, like that the class is closed under desuspension. For example, if $\mathcal C$ consists of all spectra with $\pi_n=0$ for $n<0$ and $\mathcal L$ consists of all those with $\pi_n=0$ for $n\ge 0$ then this pair is not a matched pair, although every $X$ has a universal map $X\to L$ and a universal map $C\to X$ and these form a fibration sequence.

2 deleted 2 characters in body

I was thinking of asking a similar question. This won't be an answer, but it might be a slightly useful reframing. Presumably an expert will give us a real answer.

For spectra $X$ and $Y$, let's write $X\perp Y$ if the mapping spectrum $Hom(X,Y)$ is contractible, i.e. if the cohomology $Y^\star X$ is trivial, i.e. if in the stable homotopy category there is no nontrivial map $X\to \Sigma^nY$ for any $n\in\mathbb Z$.

Consider pairs $(\mathcal C,\mathcal L)$ of classes of stable homotopy types such that ($X\in \mathcal C$ iff $X\perp Y$ for all $Y\in \mathcal L$) and ($Y\in \mathcal L$ iff $X\perp Y$ for all $X\in \mathcal C$). Let's call such a thing a matched pair, just to have a term for it.

The best matched pairs are those which have localization functors. This means that for every spectrum $X$ there is a universal example of a map $X\to L$ from $X$ to an element of $\mathcal L$.

Here are some properties that a class $C$ of stable homotopy types may or may not have:

1. $\mathcal C$ is the first member of a localization pair $(\mathcal C,\mathcal L)$.

2. $\mathcal C$ is the first member of a matched pair $(\mathcal C,\mathcal L)$.

3. There is a class $\mathcal E$ of spectra such that $X\in \mathcal C$ iff for every $E\in \mathcal E$ we have $E_\star X=0$.

4. $\mathcal C$ is closed under homotopy colimits and desuspension.

Of course 1 implies 2. The converse seems like a good question.

The question of whether a given matched pair is a localization pair can be reformulated in a couple of ways. Given a fibration sequence $\star\to C\to X\to L\to \star$, the following three conditions are equivalent:

(a) $L\in \mathcal L$ and $X\to L$ is universal among maps from $X$ to objects in $\mathcal L$.

(b) $C\in \mathcal C$ and $L\in \mathcal L$.

(c) $C\in \mathcal C$ and $C\to X$ is universal among maps to $X$ from objects in $\mathcal C$.

Clearly 2 implies 4 and 3 implies 4.

3 is like saying that $\mathcal C$ is a Bousfield acyclicity class, except that we are allowing $\mathcal E$ to be a proper class instead of a set.

(In the same sense 2 says that $\mathcal C$ is what might be called a cohomological Bousfield class.)

If we take a "What, me worry?" attitude toward set theory, then 3 implies 1: every Bousfield class has a localization functor.

In fact it seems that within the framework of this "Alfred E. von Neumann set theory" 4 implies 1, by the usual argument (carried out rigorously by Bousfield with appropriate hypotheses): To create a localization of $X$, start throwing elements of $\mathcal C$ at it and take cofibers, and keep doing this transfinitely until you are done. The resulting spectrum $L$ should belong the class $\mathcal L$ of all spectra that satisfy $C\perp L$ for all $C\in\mathcal C$. The (co)fiber of the map $X\to L$ is made from $\mathcal C$ by repeated colimits, and so is in $\mathcal C$. This makes $X\to L$ universal. Applying that same construction when $X$ satisfies $X\perp L$ for $L\in \mathcal L$, we find that such $X$ is a retract of an object in $\mathcal C$, therefore is in $\mathcal C$.

So modulo set theory 1=2=4=11=2=4.

But as far as I know the question of whether 3 is strictly stronger is of a different nature.

About (co)reflexivity: 1 implies that $\mathcal C$ is coreflexive. To get a converse you have to assume more, like that the class is closed under desuspension. For example, if $\mathcal C$ consists of all spectra with $\pi_n=0$ for $n<0$ and $\mathcal L$ consists of all those with $\pi_n=0$ for $n\ge 0$ then this pair is not a matched pair, although every $X$ has a universal map $X\to L$ and a universal map $C\to X$ and these form a fibration sequence.

1

I was thinking of asking a similar question. This won't be an answer, but it might be a slightly useful reframing. Presumably an expert will give us a real answer.

For spectra $X$ and $Y$, let's write $X\perp Y$ if the mapping spectrum $Hom(X,Y)$ is contractible, i.e. if the cohomology $Y^\star X$ is trivial, i.e. if in the stable homotopy category there is no nontrivial map $X\to \Sigma^nY$ for any $n\in\mathbb Z$.

Consider pairs $(\mathcal C,\mathcal L)$ of classes of stable homotopy types such that ($X\in \mathcal C$ iff $X\perp Y$ for all $Y\in \mathcal L$) and ($Y\in \mathcal L$ iff $X\perp Y$ for all $X\in \mathcal C$). Let's call such a thing a matched pair, just to have a term for it.

The best matched pairs are those which have localization functors. This means that for every spectrum $X$ there is a universal example of a map $X\to L$ from $X$ to an element of $\mathcal L$.

Here are some properties that a class $C$ of stable homotopy types may or may not have:

1. $\mathcal C$ is the first member of a localization pair $(\mathcal C,\mathcal L)$.

2. $\mathcal C$ is the first member of a matched pair $(\mathcal C,\mathcal L)$.

3. There is a class $\mathcal E$ of spectra such that $X\in \mathcal C$ iff for every $E\in \mathcal E$ we have $E_\star X=0$.

4. $\mathcal C$ is closed under homotopy colimits and desuspension.

Of course 1 implies 2. The converse seems like a good question.

The question of whether a given matched pair is a localization pair can be reformulated in a couple of ways. Given a fibration sequence $\star\to C\to X\to L\to \star$, the following three conditions are equivalent:

(a) $L\in \mathcal L$ and $X\to L$ is universal among maps from $X$ to objects in $\mathcal L$.

(b) $C\in \mathcal C$ and $L\in \mathcal L$.

(c) $C\in \mathcal C$ and $C\to X$ is universal among maps to $X$ from objects in $\mathcal C$.

Clearly 2 implies 4 and 3 implies 4.

3 is like saying that $\mathcal C$ is a Bousfield acyclicity class, except that we are allowing $\mathcal E$ to be a proper class instead of a set.

(In the same sense 2 says that $\mathcal C$ is what might be called a cohomological Bousfield class.)

If we take a "What, me worry?" attitude toward set theory, then 3 implies 1: every Bousfield class has a localization functor.

In fact it seems that within the framework of this "Alfred E. von Neumann set theory" 4 implies 1, by the usual argument (carried out rigorously by Bousfield with appropriate hypotheses): To create a localization of $X$, start throwing elements of $\mathcal C$ at it and take cofibers, and keep doing this transfinitely until you are done. The resulting spectrum $L$ should belong the class $\mathcal L$ of all spectra that satisfy $C\perp L$ for all $C\in\mathcal C$. The (co)fiber of the map $X\to L$ is made from $\mathcal C$ by repeated colimits, and so is in $\mathcal C$. This makes $X\to L$ universal. Applying that same construction when $X$ satisfies $X\perp L$ for $L\in \mathcal L$, we find that such $X$ is a retract of an object in $\mathcal C$, therefore is in $\mathcal C$.

So modulo set theory 1=2=4=1.

But as far as I know the question of whether 3 is strictly stronger is of a different nature.

About (co)reflexivity: 1 implies that $\mathcal C$ is coreflexive. To get a converse you have to assume more, like that the class is closed under desuspension. For example, if $\mathcal C$ consists of all spectra with $\pi_n=0$ for $n<0$ and $\mathcal L$ consists of all those with $\pi_n=0$ for $n\ge 0$ then this pair is not a matched pair, although every $X$ has a universal map $X\to L$ and a universal map $C\to X$ and these form a fibration sequence.