If $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, is $κ$ necessarily $μ$-compact?

The hypothesis is a weakening of supercompactness; it asserts that if there is any bijection from $X$ to $μ$, then at least one such bijection is in $M$. However, I doubt that it is equivalent to supercompactness. I believe it would have more relations to strong compactness (perhaps be a weakening).

On the other hand, it doesn't seem to immediately come out of the ultrapower embeddings on a measurable cardinal's ultrafilter. More specifically, in $L[U]$ (the inner model of a measurable) it holds that if $\kappa$ is measurable, then no ultrapower from an ultrafilter on a $\kappa$-sized set witnesses this large cardinal property for $\mu=\kappa^+$.

This is not to say that measurability isn't equivalent, merely that if it is equivalent to measurability, it likely would be found to be characterized by ultrafilters on sets of size above $\kappa$.

Say cardinality is $M$-downward absolute at $\mu$ when for any $X\in M$ with $|X|=\mu$, $|X|^M=\mu$.

Theorem: If $\kappa$ is measurable with $2^\kappa=\kappa^+$, and $j:V\prec M$ is an ultrapower embedding with critical point $\kappa$ from an ultrafilter on a set $S$ of size $\kappa$, then cardinality isn't $M$-downward absolute at $\kappa^+$.

Proof:

1. For every $f:S\rightarrow\kappa$, define $m(f)=[f]_U$, and it is clear that $j(\kappa)\subseteq m$$"$$\{[f]\;|\;f:S\rightarrow\kappa\}$ and so $|j(\kappa)|\leq\kappa^\kappa=\kappa^+$.
2. $\mathcal{P}^M(\kappa)$=$\mathcal{P}(\kappa)$ so $2^\kappa\leq(2^\kappa)^M$.
3. Clearly, $(2^\kappa)^M\leq 2^\kappa$, so $(2^\kappa)^M=2^\kappa=\kappa^+$.
4. Assume $M$ preserves cardinality at $\kappa^+$. Then, $|j(\kappa)|^M=\kappa^+=(2^\kappa)^M$, meaning $|j(\kappa)|^M=(2^\kappa)^M$. This is a contradiction because $j(\kappa)$ is inaccessible (in fact measurable) in $M$.

The Main Questions:

1. Is there some way to characterize the existence of a $j:V\prec M$ with critical point $\kappa$ such that cardinality is $M$-downward absolute at $\mu$ as the existence of an ultrafilter on some set?
2. The question of the title: if $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, is $κ$ necessarily $μ$-compact?
3. Does this property relate in any way to the covering property characterization of strongly compact cardinals?

More specifically, if $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, does $κ$ necessarily have some form of $μ$-compactness? Is it related to strong compactness in any way?

The hypothesis is a weakening of supercompactness; it asserts that if there is any bijection from $X$ to $μ$, then at least one such bijection is in $M$. However, I doubt that it is equivalent to supercompactness. I believe it would have more relations to strong compactness (perhaps be a weakening).

On the other hand, it doesn't seem to immediately come out of the ultrapower embeddings on a measurable cardinal's ultrafilter. More specifically, in $L[U]$ (the inner model of a measurable) it holds that if $\kappa$ is measurable, then no ultrapower from an ultrafilter on a $\kappa$-sized set witnesses this large cardinal property for $\mu=\kappa^+$.

This is not to say that measurability isn't equivalent, merely that if it is equivalent to measurability, it likely would be found to be characterized by ultrafilters on sets of size above $\kappa$.

Say cardinality is $M$-downward absolute at $\mu$ when for any $X\in M$ with $|X|=\mu$, $|X|^M=\mu$.

Theorem: If $\kappa$ is measurable with $2^\kappa=\kappa^+$, and $j:V\prec M$ is an ultrapower embedding with critical point $\kappa$ from an ultrafilter on a set $S$ of size $\kappa$, then cardinality isn't $M$-downward absolute at $\kappa^+$.

Proof:

1. For every $f:S\rightarrow\kappa$, define $m(f)=[f]_U$, and it is clear that $j(\kappa)\subseteq m$$"$$\{[f]\;|\;f:S\rightarrow\kappa\}$ and so $|j(\kappa)|\leq\kappa^\kappa=\kappa^+$.
2. $\mathcal{P}^M(\kappa)$=$\mathcal{P}(\kappa)$ so $2^\kappa\leq(2^\kappa)^M$.
3. Clearly, $(2^\kappa)^M\leq 2^\kappa$, so $(2^\kappa)^M=2^\kappa=\kappa^+$.
4. Assume $M$ preserves cardinality at $\kappa^+$. Then, $|j(\kappa)|^M=\kappa^+=(2^\kappa)^M$, meaning $|j(\kappa)|^M=(2^\kappa)^M$. This is a contradiction because $j(\kappa)$ is inaccessible (in fact measurable) in $M$.

The Main Questions:

1. If $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, is $κ$ necessarily related to some form of $μ$-compactness?
2. Is there some way to characterize the existence of a $j:V\prec M$ with critical point $\kappa$ such that cardinality is $M$-downward absolute at $\mu$ as the existence of an ultrafilter on some set?
3. Does this property relate in any way to the covering property characterization of strongly compact cardinals?

If $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, is $κ$ necessarily $μ$-compact?

The hypothesis is a weakening of supercompactness; it asserts that if there is any bijection from $X$ to $μ$, then at least one such bijection is in $M$. However, I doubt that it is equivalent to supercompactness. I believe it would have more relations to strong compactness (perhaps be a weakening).

On the other hand, it doesn't seem to immediately come out of the ultrapower embeddings on a measurable cardinal's ultrafilter. More specifically, in $L[U]$ (the inner model of a measurable) it holds that if $\kappa$ is measurable, then no ultrapower from an ultrafilter on a $\kappa$-sized set witnesses this large cardinal property for $\mu=\kappa^+$.

This is not to say that measurability isn't equivalent, merely that if it is equivalent to measurability, it likely would be found to be characterized by ultrafilters on sets of size above $\kappa$.

Say cardinality is $M$-downward absolute at $\mu$ when for any $X\in M$ with $|X|=\mu$, $|X|^M=\mu$.

Theorem: If $\kappa$ is measurable with $2^\kappa=\kappa^+$, and $j:V\prec M$ is an ultrapower embedding with critical point $\kappa$ from an ultrafilter on a set $S$ of size $\kappa$, then cardinality isn't $M$-downward absolute at $\kappa^+$.

Proof:

1. For every $f:S\rightarrow\kappa$, define $m(f)=[f]_U$, and it is clear that $j(\kappa)\subseteq m$$"$$\{[f]\;|\;f:S\rightarrow\kappa\}$ and so $|j(\kappa)|\leq\kappa^\kappa=\kappa^+$.
2. $\mathcal{P}^M(\kappa)$=$\mathcal{P}(\kappa)$ so $2^\kappa\leq(2^\kappa)^M$.
3. Clearly, $(2^\kappa)^M\leq 2^\kappa$, so $(2^\kappa)^M=2^\kappa=\kappa^+$.
4. Assume $M$ preserves cardinality at $\kappa^+$. Then, $|j(\kappa)|^M=\kappa^+=(2^\kappa)^M$, meaning $|j(\kappa)|^M=(2^\kappa)^M$. This is a contradiction because $j(\kappa)$ is inaccessible (in fact measurable) in $M$.

The Main Questions:

1. Is there some way to characterize the existence of a $j:V\prec M$ with critical point $\kappa$ such that cardinality is $M$-downward absolute at $\mu$?
2. The question of the title: if $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, is $κ$ necessarily $μ$-compact?
3. Does this property relate in any way to the covering property characterization of strongly compact cardinals?

If $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, is $κ$ necessarily $μ$-compact?

The hypothesis is a weakening of supercompactness; it asserts that if there is any bijection from $X$ to $μ$, then at least one such bijection is in $M$. However, I doubt that it is equivalent to supercompactness. I believe it would have more relations to strong compactness (perhaps be a weakening).

On the other hand, it doesn't seem to immediately come out of the ultrapower embeddings on a measurable cardinal's ultrafilter. More specifically, in $L[U]$ (the inner model of a measurable) it holds that if $\kappa$ is measurable, then no ultrapower from an ultrafilter on a $\kappa$-sized set witnesses this large cardinal property for $\mu=\kappa^+$.

This is not to say that measurability isn't equivalent, merely that if it is equivalent to measurability, it likely would be found to be characterized by ultrafilters on sets of size above $\kappa$.

Say cardinality is $M$-downward absolute at $\mu$ when for any $X\in M$ with $|X|=\mu$, $|X|^M=\mu$.

Theorem: If $\kappa$ is measurable with $2^\kappa=\kappa^+$, and $j:V\prec M$ is an ultrapower embedding with critical point $\kappa$ from an ultrafilter on a set $S$ of size $\kappa$, then cardinality isn't $M$-downward absolute at $\kappa^+$.

Proof:

1. For every $f:S\rightarrow\kappa$, define $m(f)=[f]_U$, and it is clear that $j(\kappa)\subseteq m$$"$$\{[f]\;|\;f:S\rightarrow\kappa\}$ and so $|j(\kappa)|\leq\kappa^\kappa=\kappa^+$.
2. $\mathcal{P}^M(\kappa)$=$\mathcal{P}(\kappa)$ so $2^\kappa\leq(2^\kappa)^M$.
3. Clearly, $(2^\kappa)^M\leq 2^\kappa$, so $(2^\kappa)^M=2^\kappa=\kappa^+$.
4. Assume $M$ preserves cardinality at $\kappa^+$. Then, $|j(\kappa)|^M=\kappa^+=(2^\kappa)^M$, meaning $|j(\kappa)|^M=(2^\kappa)^M$. This is a contradiction because $j(\kappa)$ is inaccessible (in fact measurable) in $M$.

The Main Questions:

1. Is there some way to characterize the existence of a $j:V\prec M$ with critical point $\kappa$ such that cardinality is $M$-downward absolute at $\mu$ as the existence of an ultrafilter on some set?
2. The question of the title: if $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, is $κ$ necessarily $μ$-compact?
3. Does this property relate in any way to the covering property characterization of strongly compact cardinals?