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I understand your question better now.

First, in your general context of filters the relations $\leq_1$ and $\leq_2$ are not the same. To see this, let $G=\{I\}$ be the trivial filter on a set $I$ with at least two points, and let $\mu$ be any nonprincipal ultrafilter on $I$. Since $G\subset \mu$, we see that $G\leq_1\mu$ as witnessed by the identity function on $I$. But for any function $f:I\to I$, it must be that $f^{-1}(j)\notin \mu$ for at least one $j$, since these are disjoint, and so $f[\mu]$ will contain $I-\{j\}$, meaning $f[\mu]\neq G$. So $G\not\leq_2\mu$, and the relations are different.

Next, I claim that for ultrafilters, the relations are the same. Suppose that $\mu\leq_1\nu$ and $\mu$ is an ultrafilter on $I$ and $\nu$ a filter on $J$. So there is a function $f:I\to J$ for which $\mu\subset f[\nu]$ in your sense. Since $f[\nu]$ is a filter and $\mu$ is an ultrafilter, this implies $\mu=f[\nu]$, and so $\mu\leq_2\nu$ as desired.

Moreover, I claim that $\leq_2$ is the same as the Rudin-Keisler order. The usual definition of this order is that if $F$ is a filter on $I$ and $f:I\to J$ is any function, then one we may define a filter $G=f*F$ on $J$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Keisler order is defined so that $G\leq_{RK} F$ if and only if there is $f$ for which $G=f*F$.

Suppose $F$ is a filter on $I$ and $f:I\to J$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset I$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset J$ shows $f*F\subset f[F]$.

It follows that $\leq_2$ is the same as the Rudin-Keisler order.

And the $\leq_1$ order is simply a combination of the subfilter relation and the Rudin-Keisler order.

I understand your question better now.

First, in your general context of filters the relations $\leq_1$ and $\leq_2$ are not the same. To see this, let $G=\{I\}$ be the trivial filter on a set $I$ with at least two points, and let $\mu$ be any nonprincipal ultrafilter on $I$. Since $G\subset \mu$, we see that $\mu\leq_1 G$ as witnessed by the identity function $i$ on $I$. (Details: since $i[I]=I$, it follows that $i[G]$ is the filter with base $\{I\}$, which is the same as $G$. So $i[G]=G$, which is a subset of $\mu$, and so $\mu\leq_1 G$.) Meanwhile, I claim that $\mu\not\leq_2 G$. To see this, observe that for any function $f:I\to I$, we have $f[G]$ is the filter with base $\{f[I]\}$, and so $f[G]\neq\mu$ since $\mu$ is nonprincipal.
So the relations are different.

Note also that if $\mu$ is an ultrafilter on $I$ and $F\leq_1 \mu$ via the function $f$ for a filter $F$, then $F$ is an ultrafilter. The reason is that if $Y\notin F$, then $f^{-1}Y\notin\mu$ and so $f^{-1}(I-Y)\in\mu$, which implies $f[f^{-1}(I-Y)]\in F$, which implies $I-Y\in F$, so $F$ is an ultrafilter.

Next, I claim that for ultrafilters, the relations are the same.

Theorem. If $\nu$ is an ultrafilter, then $F\leq_1\nu\iff F\leq_2\nu$.

Proof. It suffices to prove the forward direction. Suppose $\nu$ is an ultrafilter on a set $J$ and $F$ is a filter on $I$ and $F\leq_1\nu$ as witnessed by $f:J\to I$. So $f[\nu]\subset F$. Consider any $X\in F$. If $f^{-1}X\in\nu$, then we get $X\supset f[f^{-1}X]\in f[\nu]$ and so $X\in f[\nu]$. Otherwise, since $\nu$ is an ultrafilter, we have $f^{-1}(I-X)\in\nu$ and so $I-X\supset f[f^{-1}(I-X)]\in f[\nu]\subset F$, which would put disjoint sets in $F$, a contradiction. QED

Finally, I claim that for ultrafilters, the relation $\leq_2$ is the same as the Rudin-Keisler order. The usual definition of this order is that if $F$ is a filter on $J$ and $f:J\to I$ is any function, then one we may define a filter $G=f*F$ on $I$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Keisler order is defined so that $G\leq_{RK} F$ if and only if there is $f$ for which $G=f*F$.

Suppose $F$ is a filter on $J$ and $f:J\to I$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset J$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset I$ shows $f*F\subset f[F]$.

It follows that $\leq_2$ is the same as the Rudin-Keisler order.

I understand your question better now.

First, in your general context of filters the relations $\leq_1$ and $\leq_2$ are not the same. To see this, let $G=\{I\}$ be the trivial filter on a set $I$ with at least two points, and let $\mu$ be any nonprincipal ultrafilter on $I$. Since $G\subset \mu$, we see that $G\leq_1\mu$ as witnessed by the identity function on $I$. But for any function $f:I\to I$, it must be that $f^{-1}(j)\notin \mu$ for at least one $j$, since these are disjoint, and so $f[\mu]$ will contain $I-\{j\}$, meaning $f[\mu]\neq G$. So $G\not\leq_2\mu$, and the relations are different.

Next, I claim that for ultrafilters, the relations are the same. Suppose that $\mu\leq_1\nu$ and $\mu$ is an ultrafilter on $I$ and $\nu$ a filter on $J$. So there is a function $f:I\to J$ for which $\mu\subset f[\nu]$ in your sense. Since $f[\nu]$ is a filter and $\mu$ is an ultrafilter, this implies $\mu=f[\nu]$, and so $\mu\leq_2\nu$ as desired.

Moreover, I claim that $\leq_2$ is the same as the Rudin-Kiesler order. The usual definition of this order is that if $F$ is a filter on $I$ and $f:I\to J$ is any function, then one we may define a filter $G=f*F$ on $J$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Kielser order is defined so that $G\leq_{RK} F$ if and only if there is $f$ for which $G=f*F$.

Suppose $F$ is a filter on $I$ and $f:I\to J$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset I$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset J$ shows $f*F\subset f[F]$.

It follows that $\leq_2$ is the same as the Rudin-Kiesler order.

And the $\leq_1$ order is simply a combination of the subfilter relation and the Rudin-Kiesler order.

I understand your question better now.

First, in your general context of filters the relations $\leq_1$ and $\leq_2$ are not the same. To see this, let $G=\{I\}$ be the trivial filter on a set $I$ with at least two points, and let $\mu$ be any nonprincipal ultrafilter on $I$. Since $G\subset \mu$, we see that $G\leq_1\mu$ as witnessed by the identity function on $I$. But for any function $f:I\to I$, it must be that $f^{-1}(j)\notin \mu$ for at least one $j$, since these are disjoint, and so $f[\mu]$ will contain $I-\{j\}$, meaning $f[\mu]\neq G$. So $G\not\leq_2\mu$, and the relations are different.

Next, I claim that for ultrafilters, the relations are the same. Suppose that $\mu\leq_1\nu$ and $\mu$ is an ultrafilter on $I$ and $\nu$ a filter on $J$. So there is a function $f:I\to J$ for which $\mu\subset f[\nu]$ in your sense. Since $f[\nu]$ is a filter and $\mu$ is an ultrafilter, this implies $\mu=f[\nu]$, and so $\mu\leq_2\nu$ as desired.

Moreover, I claim that $\leq_2$ is the same as the Rudin-Keisler order. The usual definition of this order is that if $F$ is a filter on $I$ and $f:I\to J$ is any function, then one we may define a filter $G=f*F$ on $J$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Keisler order is defined so that $G\leq_{RK} F$ if and only if there is $f$ for which $G=f*F$.

Suppose $F$ is a filter on $I$ and $f:I\to J$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset I$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset J$ shows $f*F\subset f[F]$.

It follows that $\leq_2$ is the same as the Rudin-Keisler order.

And the $\leq_1$ order is simply a combination of the subfilter relation and the Rudin-Keisler order.

I understand your question better now.

First, in your general context of filters the relations $\leq_1$ and $\leq_2$ are not the same. To see this, let $G=\{I\}$ be the trivial filter on a set $I$ with at least two points, and let $\mu$ be any nonprincipal ultrafilter on $I$. Since $G\subset \mu$, we see that $G\leq_1\mu$ as witnessed by the identity function on $I$. But for any function $f:I\to I$, it must be that $f^{-1}(j)\notin \mu$ for at least one $j$, since these are disjoint, and so $f[\mu]$ will contain $I-\{j\}$, meaning $f[\mu]\neq G$. So $G\not\leq_2\mu$, and the relations are different.

Next, I claim that for ultrafilters, the relations are the same. Suppose that $\mu\leq_1\nu$ and $\mu$ is an ultrafilter on $I$ and $\nu$ a filter on $J$. So there is a function $f:I\to J$ for which $\mu\of f[\nu]$ in your sense. Since $f[\nu]$ is a filter and $\mu$ is an ultrafilter, this implies $\mu=f[\nu]$, and so $\mu\leq_2\nu$ as desired.

Moreover, I claim that $\leq_2$ is the same as the Rudin-Kiesler order. The usual definition of this order is that if $F$ is a filter on $I$ and $f:I\to J$ is any function, then one we may define a filter $f*F$ on $J$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Kielser order is defined so that $G\leq_{RK] F$ if and only if there is $f$ for which $G=f*F$.

Suppose $F$ is a filter on $I$ and $f:I\to J$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset I$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset J$ shows $f*F\subset f[F]$.

It follows that $\leq_2$ is the same as the Rudin-Kiesler order.

And the $\leq_1$ order is simply a combination of the subfilter relation and the Rudin-Kiesler order.

I understand your question better now.

First, in your general context of filters the relations $\leq_1$ and $\leq_2$ are not the same. To see this, let $G=\{I\}$ be the trivial filter on a set $I$ with at least two points, and let $\mu$ be any nonprincipal ultrafilter on $I$. Since $G\subset \mu$, we see that $G\leq_1\mu$ as witnessed by the identity function on $I$. But for any function $f:I\to I$, it must be that $f^{-1}(j)\notin \mu$ for at least one $j$, since these are disjoint, and so $f[\mu]$ will contain $I-\{j\}$, meaning $f[\mu]\neq G$. So $G\not\leq_2\mu$, and the relations are different.

Next, I claim that for ultrafilters, the relations are the same. Suppose that $\mu\leq_1\nu$ and $\mu$ is an ultrafilter on $I$ and $\nu$ a filter on $J$. So there is a function $f:I\to J$ for which $\mu\subset f[\nu]$ in your sense. Since $f[\nu]$ is a filter and $\mu$ is an ultrafilter, this implies $\mu=f[\nu]$, and so $\mu\leq_2\nu$ as desired.

Moreover, I claim that $\leq_2$ is the same as the Rudin-Kiesler order. The usual definition of this order is that if $F$ is a filter on $I$ and $f:I\to J$ is any function, then one we may define a filter $G=f*F$ on $J$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Kielser order is defined so that $G\leq_{RK} F$ if and only if there is $f$ for which $G=f*F$.

Suppose $F$ is a filter on $I$ and $f:I\to J$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset I$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset J$ shows $f*F\subset f[F]$.

It follows that $\leq_2$ is the same as the Rudin-Kiesler order.

And the $\leq_1$ order is simply a combination of the subfilter relation and the Rudin-Kiesler order.