We're going to need to go out of order here a bit and do Proposition 31 before doing Propositions 29 and 30. Please have patience.

Proposition 31: All five characterizations of the pullback define the same infradistribution.

So, the two requirements of being able to do a pullback $g^{\ast }\left(h\right)$ are that: $h$ has $g\left(X\right)$ as a support, and $g$ is a proper function (preimage of a compact set is compact). And our five characterizations are:

1: For sets,

$g^{\ast }\left(H\right):=\left(g_{\ast }{)}^{-1}\right(H)$

ie, take the preimage of $H$ w.r.t. the linear operator that $g$ induces from $M^a\left(X\right)\to M^a\left(Y\right)$.

2: For concave functionals, $g^{\ast }\left(h\right)$ is the concave monotone hull of the partial function $f\circ g\mapsto h\left(f\right)$.

3: For concave functionals, $g^{\ast }\left(h\right)$ is the monotone hull of the partial function $f\circ g\mapsto h\left(f\right)$, ie:

$g^{\ast }\left(h\right)\left(f\right):={sup}_{f^{\ast }:f^{\ast }\circ g\le f}h\left(f^{\ast }\right)$

4: For concave functionals, 

$g^{\ast }\left(h\right)=inf\{h^{\prime }|h^{\prime }\in \square X,g_{\ast }(h^{\prime })=h\}$

take the inf of all infradistributions that could have made your infradistribution of interest via pushforward.
 

5: For sets, take the closed convex hull of all infradistribution sets which project down to be equivalent to $H$.
 

The proof path proof path will be first establishing that 3 makes an infradistribution, then showing that 3 and 2 are equivalent, then showing that 2 and 4 are equivalent, then showing that 4 and 5 are equivalent. And now that 2, 3, 4, and 5 are in an equivalence class, we wrap up by showing that 2 implies 1 and 1 implies 5.

First up: Establishing that 3 makes an infradistribution, which is where the bulk of the work is.

Starting with normalization:

$g^{\ast }\left(h\right)\left(1\right):={sup}_{f^{\ast }:f^{\ast }\circ g\le 1}h\left(f^{\ast }\right)$

And clearly, this supremum is attained by $f^{\ast }=1$, it can be no higher, by normalization, so we have $h\left(1\right)=1$ and thus one part of normalization. For the other half of normalization,

$g^{\ast }\left(h\right)\left(0\right)={sup}_{f^{\ast }:f^{\ast }\circ g\le 0}h\left(f^{\ast }\right)$

Now, in order for $f^{\ast }\circ g\le 0$ to be the case, all we need is that $f^{\ast }$ be 0 on $g\left(X\right)$. However, since $g\left(X\right)$ is assumed to be a support for $h$, all functions like this must have $h\left(f^{\ast }\right)=h\left(0\right)=0$ and we have the other part of normalization.

Monotonicity: This is immediate because if $f^{\prime }\ge f$, then the supremum for $f^{\prime }$ has more options and can attain a higher value.

Concavity:

$p{g}^{\ast }\left(h\right)\left(f\right)+(1-p)g^{\ast }\left(h\right)\left(f^{\prime }\right)=p{sup}_{f^{\ast }:f^{\ast }\circ g\le 1}h\left(f^{\ast }\right)+(1-p){sup}_{f^{{}^{\prime }\ast }:f^{{}^{\prime }\ast }\circ g\le 1}h\left(f^{{}^{\prime }\ast }\right)$

And we can pick some $f^{\ast }$ and $f^{{}^{\prime }\ast }$ which very nearly attain the relevant values, as close as you wish, to go:

$\simeq ph\left(f^{\ast }\right)+(1-p)h\left(f^{{}^{\prime }\ast }\right)\le h(p{f}^{\ast }+(1-p\left)f^{{}^{\prime }\ast }\right)$

And then,

$(p{f}^{\ast }+(1-p\left)f^{{}^{\prime }\ast }\right)\circ g=p(f^{\ast }\circ g)+(1-p)(f^{{}^{\prime }\ast }\circ g)\le pf+(1-p)f^{\prime }$

So, with that, we can go:

$\le {sup}_{f^{\ast }:f^{\ast }\circ g\le pf+(1-p)f^{\prime }}h\left(f^{\ast }\right)=g^{\ast }\left(h\right)(pf+(1-p\left)f^{\prime }\right)$

And we're done with concavity.

For Lipschitzness, the proof will work as follows. We'll take two functions $f$ and $f^{\prime }$, and show that $g^{\ast }\left(h\right)\left(f^{\prime }\right)\ge g^{\ast }\left(h\right)\left(f\right)-{\lambda }^{\odot }d(f,f^{\prime })$ in a way that generalizes to symmetrically show that $g^{\ast }\left(h\right)\left(f\right)\ge g^{\ast }\left(h\right)\left(f^{\prime }\right)-{\lambda }^{\odot }d(f,f^{\prime })$ to get our result that

$|g^{\ast }\left(h\right)\left(f\right)-g^{\ast }\left(h\right)\left(f^{\prime }\right)|\le {\lambda }^{\odot }d(f,f^{\prime })$

where ${\lambda }^{\odot }$ is the Lipschitz constant of $h$.

So, let's show

$g^{\ast }\left(h\right)\left(f^{\prime }\right)\ge g^{\ast }\left(h\right)\left(f\right)-{\lambda }^{\odot }d(f,f^{\prime })$

What we can do is, because

$g^{\ast }\left(h\right)\left(f\right)={sup}_{f^{\ast }:f^{\ast }\circ g\le f}h\left(f^{\ast }\right)$

then we can consider this value to be approximately obtained by some specific function $f^{\ast }$. Then,

$g^{\ast }\left(h\right)\left(f^{\prime }\right)\ge g^{\ast }\left(h\right)(inf(f,f^{\prime }\left)\right)\ge g^{\ast }\left(h\right)(f-d(f,f^{\prime }\left)\right)\ge h(f^{\ast }-d(f,f^{\prime }\left)\right)$

$\ge h\left(f^{\ast }\right)-{\lambda }^{\odot }d(f,f^{\prime })\simeq g^{\ast }\left(h\right)\left(f\right)-{\lambda }^{\odot }d(f,f^{\prime })$

The first two inequalities are from monotonicity. The third inequality is because $(f^{\ast }-d(f,f^{\prime }\left)\right)\circ g\le f-d(f,f^{\prime }))$, where $f^{\ast }$ is our special function from a bit earlier, because $f^{\ast }\circ g\le f$. So, it isn't necessarily the supremum and doesn't attain maximum value. Then, there's just Lipschitzness of $h$, and $h\left(f^{\ast }\right)$ being approximately $g^{\ast }\left(h\right)\left(f\right)$. And we're done with Lipschitzness, just swap $f$ and $f^{\prime }$ and use earlier arguments.

Before proceeding further, note the following information: Polish spaces are first-countable and Hausdorff so they're compactly generated, and Wikipedia says any proper map to a compactly generated space is closed (maps closed sets to closed sets). So, $g\left(X\right)$ is a closed set.

Now just leaves compact almost-support. Fix an $ϵ$, and take a compact $ϵ$-almost-support of $h$, $C_{ϵ}\subseteq Y$. Intersect with $g\left(X\right)$ if you haven't already, it still makes a compact set and it keeps being an $ϵ$-almost-support. We will show that $g^{-1}\left(C_{ϵ}\right)$, which is compact because we assumed our map was proper, is an $ϵ$-almost-support of $g^{\ast }\left(h\right)$.

This proof will proceed by showing that, if $f,f^{\prime }$ agree on $g^{-1}\left(C_{ϵ}\right)$, then $g^{\ast }\left(h\right)\left(f\right)-ϵd(f,f^{\prime })\le g^{\ast }\left(h\right)\left(f^{\prime }\right)$. Symmetrical arguments work by swapping $f$ and $f^{\prime }$, showing that the two expectation values are only $ϵd(f,f^{\prime })$ apart.

As usual, we have

$g^{\ast }\left(h\right)\left(f\right)\simeq h\left(f^{\ast }\right)$

Where $f^{\ast }$ very nearly attains the supremum, and we know that $f^{\ast }\circ g\le f$.

Let the function $\Theta :g\left(X\right)\to P\left(R\right)$ (a set-valued function) be defined as:

If $y\in C_{ϵ}$, then $\Theta \left(y\right)=f^{\ast }\left(y\right)$

If $y\notin C_{ϵ}$, then

$\Theta \left(y\right):=\left[f^{\ast }\right(y)-d(f,f^{\prime }),inf({min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right),f^{\ast }\left(y\right)+d(f,f^{\prime })\left)\right]$

These sets are always concave, and also always nonempty, because, in the second case, the only way it could be nonempty is if

${min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right)<f^{\ast }\left(y\right)-d(f,f^{\prime })$

Now, g being a proper map, the preimage of the single point $y$ is a compact set, so we can actually minimize our function, pick out our particular $x$ that maps to $y$. Then, we know:

$f^{\prime }\left(x\right)+d(f,f^{\prime })\ge f\left(x\right)\ge f^{\ast }\left(g\right(x\left)\right)=f^{\ast }\left(y\right)$

The second inequality is because $f\ge f^{\ast }\circ g$ by assumption. Shuffling this around a bit, we have:

$f^{\prime }\left(x\right)\ge f^{\ast }\left(y\right)-d(f,f^{\prime })$

where our $x$ is minimizing, which is incompatible with nonemptiness.

So, we've got convexity and nonemptiness. Let's shower lower-hemicontinuity.

As a warmup for that, we'll need to show that the function $y\mapsto {min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right)$
Is lower-semicontinuous.

So, our proof goal is: if $y_n$ limits to $y$, then

${liminf}_{n\to \infty }{min}_{x\in g^{-1}\left(y_n\right)}{f}^{\prime }\left(x\right)\ge {min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right)$

And we know that all our $y_n$ and $y$ itself, lie in $g\left(X\right)$.

At this point, here's what we do. Isolate the subsequence of points we're using to attain the $liminf$. This sequence of points $y_n$ (when unioned with $y$ itself) is compact in $Y$, so its preimage in $X$ is compact. And you can minimize functions over compact sets. So, for each $y_n$, pick a point $x_n\in g^{-1}\left(y_n\right)$ that minimizes $f^{\prime }$. Our sequence of points $x_n$ is roaming around in a compact set, so we can pick a convergent subsequence. At this point, let's start indexing with $m$. Said limit point $x$ must have $f^{\prime }\left(x\right)={lim}_{m\to \infty }{f}^{\prime }\left(x_m\right)$ by continuity for $f^{\prime }$, and $g\left(x\right)=y$ by continuity for $g$ (all the $x_m$ get mapped to $y_m$, so the limit point $x$ must go to the limit point $y$) Putting it all together,

${liminf}_{n\to \infty }{min}_{x\in g^{-1}\left(y_n\right)}{f}^{\prime }\left(x\right)={lim}_{m\to \infty }{f}^{\prime }\left(x_m\right)=f^{\prime }\left(x\right)\ge {inf}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right)$

And so we have that, if $y_n$ limits to $y$, then

${liminf}_{n\to \infty }{min}_{x\in g^{-1}\left(y_n\right)}{f}^{\prime }\left(x\right)\ge {min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right)$

And $y\mapsto {min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right)$ is lower-semicontinuous over $g\left(X\right)$.

At this point, we'll try to show that the set-valued function $\Theta$ is lower-hemicontinuous over $g\left(X\right)$. We'll use z to denote numbers in $R$.

The criterion for lower-hemicontinuity of $\Theta$ is: If $y_n$ limits to $y$, and $z\in \Theta \left(y\right)$, then there's a subsequence $y_m$ and points $z_m\in \Theta \left(y_m\right)$ where ${lim}_{m\to \infty }{z}_m=z$

To show this, we'll divide into three cases. In our first case, only finitely many of the $y_n$ are in $C_{ϵ}$. Let's pass to a subsequence where we remove all the $y_n$ that lie in $C_{ϵ}$, and where ${min}_{x\in g^{-1}\left(y_m\right)}{f}^{\prime }\left(x\right)$ converges to a value. Let 

$z_m:=sup(inf(z,{min}_{x\in g^{-1}\left(y_m\right)}{f}^{\prime }\left(x\right),f^{\ast }\left(y_m\right)+d(f,f^{\prime })),f^{\ast }(y_m)-d(f,f^{\prime }\left)\right)$

Then,

$z=sup(inf(z,{min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right),f^{\ast }\left(y\right)+d(f,f^{\prime })),f^{\ast }(y)-d(f,f^{\prime }\left)\right)$

$=sup(inf(z,{lim}_{m\to \infty }{min}_{x\in g^{-1}\left(y_m\right)}{f}^{\prime }\left(x\right),{lim}_{m\to \infty }{f}^{\ast }\left(y_m\right)+d(f,f^{\prime })\left)\right),{lim}_{m\to \infty }{f}^{\ast }\left(y_m\right)-d(f,f^{\prime }))$

$=sup({lim}_{m\to \infty }inf(z,{min}_{x\in g^{-1}\left(y_m\right)}{f}^{\prime }\left(x\right),f^{\ast }\left(y_m\right)+d(f,f^{\prime })),{lim}_{m\to \infty }{f}^{\ast }(y_m)-d(f,f^{\prime }\left)\right)$

$={lim}_{m\to \infty }sup(inf(z,{min}_{x\in g^{-1}\left(y_m\right)}{f}^{\prime }\left(x\right),f^{\ast }\left(y_m\right)+d(f,f^{\prime })),f^{\ast }(y_m)-d(f,f^{\prime }\left)\right)={lim}_{m\to \infty }{z}_m$

This was because all these sequences have their corresponding values converge, whether by continuity or assumption. So, this takes care of lower-hemicontinuity in one case.

Now for our second case, where only finitely may of the $y_n$ aren't in $C_{ϵ}$. Pass to a subsequence where they are all removed, and where ${min}_{x\in g^{-1}\left(y_m\right)}{f}^{\prime }\left(x\right)$ converges to a value. Accordingly, all our  $y_m$ lie in $C_{ϵ}$. Let $z_m=f^{\ast }\left(y_m\right)$. Then

${lim}_{m\to \infty }{z}_m={lim}_{m\to \infty }{f}^{\ast }\left(y_m\right)=f^{\ast }\left(y\right)=z$

Our third case is the trickiest, where there are infinitely many $y_n$ in $C_{ϵ}$, and infinitely many outside that set, and accordingly, $y\in C_{ϵ}$ Let's pass to a subsequence where we remove all the $y_n$ that lie in $C_{ϵ}$ and where ${min}_{x\in g^{-1}\left(y_m\right)}{f}^{\prime }\left(x\right)$ converges to a value. We would be able to apply the proof from the first case if we knew that

$f^{\ast }\left(y\right)\in \left[f^{\ast }\right(y)-d(f,f^{\prime }),inf({min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right),f^{\ast }\left(y\right)+d(f,f^{\prime })\left)\right]$

Obviously, we have $f^{\ast }\left(y\right)\ge f^{\ast }\left(y\right)-d(f,f^{\prime })$, but we still need to show that

$f^{\ast }\left(y\right)\le {min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right)$

and

$f^{\ast }\left(y\right)\le f^{\ast }\left(y\right)+d(f,f^{\prime })$

The second one can be addressed trivially, so now we just have to show that

$f^{\ast }\left(y\right)\le {min}_{x\in g^{-1}\left(y\right)}{f}^{\prime }\left(x\right)$

Pick some minimizing $x$, where $g\left(x\right)=y$, and remember that this $y$ lies in $C_{ϵ}$, so $x\in g^{-1}\left(C_{ϵ}\right)$, the compact set where $f=f^{\prime }$. Now we can just go:

$f^{\ast }\left(y\right)=f^{\ast }\left(g\right(x\left)\right)\le f\left(x\right)=f^{\prime }\left(x\right)$

And we're done. That's the last piece we need to know that $\Theta$ is lower-hemicontinuous.

Now we will invoke the Michael selection theorem, which gets us a continuous selection $f^{{\ast }^{\prime }}:Y\to R$ (bounded because $f$ and $f^{\prime }$ are) where, for all $y$, $f^{{\ast }^{\prime }}\left(y\right)\in \Theta \left(y\right)$.

The conditions to invoke the Michael selection theorem are that $F$ is lower-hemicontinuous (check), that $Y$ is paracompact, and that $R$ is a Banach space (check). Fortunately, any metrizable space is paracompact, and Polish spaces are metrizable, so we can invoke the theorem and get our continuous $f^{{\ast }^{\prime }}$.

Note two important things about this continuous $f^{{\ast }^{\prime }}$ we just constructed. First, it perfectly matches $f^{\ast }$ over $C_{ϵ}$. Second, for all $x$ in $g\left(X\right)$ (remember, that's the space it's defined over)

$(f^{{\ast }^{\prime }}\circ g)\left(x\right)=f^{{\ast }^{\prime }}\left(g\right(x\left)\right)\le {min}_{x^{\prime }\in g^{-1}\left(g\right(x\left)\right)}{f}^{\prime }\left(x^{\prime }\right)\le f^{\prime }\left(x\right)$

Because of that worst-case bound thing being one of the upper bounds on $\Theta \left(y\right)$. Also, over $g\left(X\right)$, it remains within $d(f,f^{\prime })$ of $f^{\ast }$, again, because of the imposed bounds.

We can use the Tietze extension theorem to extend $f^{{\ast }^{\prime }}$ to all of $X$ while staying within $d(f,f^{\prime })$ of $f^{\ast }$.

At this point, we can return to where we started out in this section, and go:

$g^{\ast }\left(h\right)\left(f\right)-ϵd(f,f^{\prime })={sup}_{f^{\ast }:f^{\ast }\circ g\le f}h\left(f^{\ast }\right)-ϵd(f,f^{\prime })\simeq h\left(f^{\ast }\right)-ϵd(f,f^{\prime })$

$\le \left(h\right(f^{{}^{\prime }\ast })+ϵd(f,f^{\prime }\left)\right)-ϵd(f,f^{\prime })=h\left(f^{{}^{\prime }\ast }\right)$

$\le {sup}_{f^{{}^{\prime }\ast }:f^{{}^{\prime }\ast }\circ g\le f^{\prime }}h\left(f^{{}^{\prime }\ast }\right)=g^{\ast }\left(h\right)\left(f^{\prime }\right)$

The first equality was the definition of the preimage, the approximate equality was because $f^{\ast }$ is very very near attaining the true supremum, the $\le$ is because $f^{{}^{\prime }\ast }$ and $f$ agree on $C_{ϵ}$ which is an $ϵ$-almost-support for $h$, the next inequality after that is because we just showered that $f^{{\ast }^{\prime }}\circ g\le f^{\prime }$, and then packing up the definition of preimage.

Perfectly symmetric arguments work when you swap $f^{\prime }$ and $f$, yielding our net result that

$|g^{\ast }\left(h\right)\left(f\right)-g^{\ast }\left(h\right)\left(f^{\prime }\right)|\le ϵd(f,f^{\prime })$

And $f,f^{\prime }$ were selected to be arbitrary continuous functions that agree on $g^{-1}\left(C_{ϵ}\right)$, which is compact, so we have a compact $ϵ$-almost-support, so $g^{\ast }\left(h\right)$ has compact almost-support and is an infradistribution.

Ok, so we're done, we've defined a pullback, it's the monotone hull of the function $h$ under precomposition with $g$, and this was definition 3. Now it's time to start showing equivalences.

First, the equality of definitions 2 and 3. The concave monotone hull of the partially defined pullback must be $\ge$ than the monotone hull of the partially defined pullback, because we're imposing an additional condition, there are less functions that fit the bill. But, we've showed that the monotone hull is concave, so it's actually an equality.

Now for the equality of of 2 and 4, that the concave monotone hull of the pullback is the same as the inf of all infradistributions that produce $h$ when pushfoward is applied.

We know that the concave monotone hull is indeed an infradistribution, and 

$g_{\ast }\left(g^{\ast }\right(h\left)\right)\left(f\right)=g^{\ast }\left(h\right)(f\circ g)={sup}_{f^{\ast }:f^{\ast }\circ g\le f\circ g}h\left(f^{\ast }\right)=h\left(f\right)$

So, since its projection forward matches $h$ perfectly, we have

$g^{\ast }\left(h\right)\ge inf\{h^{\prime }\in \square X|g_{\ast }(h^{\prime })=h\}$

For actual equality, we note that, if $g_{\ast }\left(h^{\prime }\right)=h$, then

$h^{\prime }(f\circ g)=g_{\ast }\left(h^{\prime }\right)\left(f\right)=h\left(f\right)$

And since $g^{\ast }\left(h\right)$ is the concave monotone hull of the partial function $f\circ g\mapsto h\left(f\right)$, and the concave monotone hull is the least concave monotone function compatible with said partial function, and any $h^{\prime }$ which pushes forward to map to $h$ must be concave and monotone and match this partial function, all the $h^{\prime }$ we're inf-ing together lie above $g^{\ast }\left(h\right)$, so we have actual equality and

$g^{\ast }\left(h\right)=inf\left\{h^{\prime }|g_{\ast }\right(h^{\prime })=h\}$

So, at this point, definitions 2, 3, and 4 of the concave hull are equal.

Now for equality of definitions 4 and 5. Taking the inf of infradistributions on the concave functional level is the same as unioning together all the infradistribution sets, and taking the closed concave hull (because that doesn't affect the expectation values), so "inf of infradistributions that pushfoward to make $h$", on the concave functional side of LF-duality, is the same as "closed concave hull and upper completion of infradistributions which pushforward to make $H$" on the set side of LF-duality.

Now that we know definitions 2-5 of the pullback are all equal, and using $h_3,h_1,h_5$ for the expectation functionals corresponding to definitions 3, 1, and 5 of the pullback, we'll show $h_5\ge h_1\ge h_3$, so $h_1$, ie, the first definition of the pullback, must be equal to the rest.

First, $h_1$ is the functional corresponding to the set:

$g^{\ast }\left(H\right):=\left(g_{\ast }{)}^{-1}\right(H)$

Where $g_{\ast }$ in this case is the function of type signature $M^a\left(X\right)\to M^a\left(Y\right)$ given by keeping the $b$ term the same and pushing forward the signed measure component via the function $g$. Note that said pushfoward can actually only make points in $M^a\left(g\right(X\left)\right)$, the measures can only be supported over that set.

We'll show that 

$E_{g^{\ast }\left(H\right)}:C_B\left(X\right)\to R$

is monotone and that 

$E_{g^{\ast }\left(H\right)}(f\circ g)=E_H\left(f\right)$

To begin with, for monotonicity, if $f^{\prime }\ge f$, then

$E_{g^{\ast }\left(H\right)}\left(f^{\prime }\right)={inf}_{(m,b)\in \left(g_{\ast }{)}^{-1}\right(H)}m\left(f^{\prime }\right)+b$

$\ge {inf}_{(m,b)\in \left(g_{\ast }{)}^{-1}\right(H)}m\left(f\right)+b$

$=E_{g^{\ast }\left(H\right)}\left(f\right)$

This is easy because our set is entirely of a-measures. So the corresponding $h_1$ is monotone.

Now, at this point, we can go:

$E_{g^{\ast }\left(H\right)}(f\circ g)={inf}_{(m,b)\in \left(g_{\ast }{)}^{-1}\right(H)}m(f\circ g)+b$

$={inf}_{(m,b):\left(g_{\ast }\right(m),b)\in H}m(f\circ g)+b$

The next line is because $g_{\ast }\left(m\right)\left(f\right)$ (the continuous pushfoward of a measure, evaluated via $f$) is the same as $m(f\circ g)$ (old measure evaluating $f\circ g$)

$={inf}_{(m,b):\left(g_{\ast }\right(m),b)\in H}{g}_{\ast }\left(m\right)\left(f\right)+b$

The next line is a bit tricky, because pushforward seems like it may only net some of the points in H. However, any measure supported over $g\left(X\right)$ has a measure over $X$ that, when pushfoward is applied, produces it. Just take each point $y\in g\left(X\right)$, craft a Markov Kernel from $g\left(X\right)$ to $X$ that maps all $y\in g\left(X\right)$ to a probability distribution supported over $g^{-1}\left(y\right)$, take the semidirect product of your measure over $g\left(X\right)$ and your kernel, and project to $X$ to get such a measure. So, said pushfoward surjects onto $H$, and we can get:

$={inf}_{(m^{\prime },b^{\prime })\in H}{m}^{\prime }\left(f\right)+b^{\prime }=E_H\left(f\right)$

So, now that we've shown that

$E_{g^{\ast }\left(H\right)}:C_B\left(X\right)\to R$

is monotone and that 

$E_{g^{\ast }\left(H\right)}(f\circ g)=E_H\left(f\right)$

We know that the corresponding functional $h_1$ has the property that it's monotone and $h_1(f\circ g)=h\left(f\right)$. Thus, it must lie above the monontone hull of $f\circ g\mapsto h\left(f\right)$, which is just $h_3$ (third definition of the pullback) and we have: $h_1\ge h_3$.

Now for the other direction, showing that $h_5\ge h_1$. The fifth definition of the pullback was the closed convex hull (and upper completion) of all the infradistributions which project to $H$. So, if we can show that any infradistribution set $H^{\prime }$ which projects into $H$ must have its minimal points lie in $g^{\ast }\left(H\right)$, we'll have that all infradistribution which project to $H$ are a subset of $g^{\ast }\left(H\right)$, and thus is higher in the ordering on functionals, $h_5\ge h_1$.

This is trivial by $g^{\ast }\left(H\right)$ just being a preimage, so any set which pushforwards into the set of interest must be a subset of the preimage.

And we have our desired result that any infradistribution set which projects into $H$ must be a subset of $g^{\ast }\left(H\right)$ (all minimal points are present in it), and higher in the information ordering, so the inf of all of them, ie, the functional $h_5$, must lie above the corresponding functional $h_1$.

So, we have $h_5\ge h_1\ge h_3=h_5$ and so $h_1$ equals all of them, and the restricted preimage of H is another formulation of the pullback, and we're done.

Proposition 29: The pullback of an infradistribution is an infradistribution, and it preserves all properties indicated in the table for this section.

So, we know from the proof of Proposition 31 that the pullback of an infradistribution is an infradistribution. Now, to show preservation of properties, we'll work with the formulation of the pullback as:

$g^{\ast }\left(H\right)=\left(g_{\ast }{)}^{-1}\right(H)$

For homogenity, 1-Lipschitzness, cohomogenity, C-additivity, and crispness, they are all characterized by all minimal points having certain properties of their $\lambda$ and $b$ values. All minimal points of $g^{\ast }\left(H\right)$ must pushforward to minimal points of $H$ (otherwise the $b$ term of your original minimal point could be strictly decreased and it'd still pushforward into $H$), which have the property of interest, and because pushforward $g_{\ast }$ preserves the $\lambda$ and $b$ values of a-measures, we know that all minimal points of $g^{\ast }\left(H\right)$ have the same property on their $\lambda$ and $b$, so all these properties are preserved under pullback.

Sharpness takes a slightly more detailed argument. Let's say that $H$ is sharp, ie, its minimal points consist entirely of the probability distributions on $C$. We know from crispness preservation that $g^{\ast }\left(H\right)$ has its minimal points consist entirely of probability distributions. If there was a probability distribution that wasn't supported on $g^{-1}\left(C\right)$ (preimage of $C$, which, by the assumed properness of $g$, is compact), that was present as a minimal point, then it would pushfoward to make a minimal point in $H$ that was a probability distribution that had some measure outside of $C$, which is impossible. Further, any probability distribution that is supported on $g^{-1}\left(C\right)$ would pushfoward to make a probability distribution supported on $C$, which would be a minimal point of $H$, so said point must be minimal in $g^{\ast }\left(H\right)$. Thus, the minimal points of $g^{\ast }\left(H\right)$ consist entirely of probability distributions supported on $g^{-1}\left(C\right)$, so $g^{\ast }\left(h\right)$ is sharp, and all nice properties are therefore preserved.

Proposition 30: Pullback then pushforward is identity, $g_{\ast }\left(g^{\ast }\right(h\left)\right)=h$, and pushforward then pullback is below identity, $g^{\ast }\left(g_{\ast }\right(h\left)\right)\le h$

The first part is easy to show,

$g_{\ast }\left(g^{\ast }\right(h\left)\right)\left(f\right)=g^{\ast }\left(h\right)(f\circ g)={sup}_{f^{\ast }:f^{\ast }\circ g\le f\circ g}h\left(f^{\ast }\right)=h\left(f\right)$

Yielding pullback then pushforward being identity. For the reverse direction, we use Proposition 31, in particular the part about the pullback being the least infradistribution that pushes forward to produce the target. And we have $g_{\ast }\left(g^{\ast }\right(g_{\ast }\left(h\right)\left)\right)=g_{\ast }\left(h\right)$, therefore $g^{\ast }\left(g_{\ast }\right(h\left)\right))\le h$

Proposition 32: The free product of two infradistributions exists and is an infradistribution iff:
1: There are points $(\lambda \mu ,0)\in H_1^{min}$ and $({\lambda }^{\prime }{\mu }^{\prime },0)\in H_2^{min}$ where $\lambda ={\lambda }^{\prime }$
2: There are points $(\lambda \mu ,b)\in H_1^{min}$ and $({\lambda }^{\prime }{\mu }^{\prime },b^{\prime })\in H_2^{min}$ where $\lambda ={\lambda }^{\prime }$, and $b=b^{\prime }$, and $\lambda +b=1$.

So, to show this, what we'll be doing is using our set-form characterization of the pullback as a preimage and the supremum as an intersection. We already know that, from proposition 14, a necessary condition for the intersection of infradistributions being an infradistribution is normalization, ie, on the set level, the presence of a point with $b=0$ and a point with $\lambda +b=1$. The presence of a point with $b=0$ in the intersection of the preimages would pushforward to make a $b=0$ point in $H_1$ and $H_2$, respectively, with equal $\lambda$ values, fulfilling condition 1.

Also, the presence of a point with $\lambda +b=1$ in the intersection of the preimages would pushforward to make a $\lambda +b=1$ minimal point in $H_1$ and $H_2$ respectively, with equal $\lambda$ values and $b$ values, fufilling condition 2.

Thus, if the free product exists and is an infradistribution, the two properties are fulfilled. That just leaves the reverse direction of establishing that the free product exists and is an infradistribution if both properties are fulfilled. 

We'll be handling the set conditions on an infradistribution. For nonemptiness and normalization, observe that if condition 1 is fulfilled, you can make the a-measure $\left(\lambda \right(\mu \times {\mu }^{\prime }),0)$ which projects down to $(\lambda \mu ,0)$ and $(\lambda {\mu }^{\prime },0)$, so it's present in the intersection of the preimages of $H_1$ and $H_2$ under the projection mappings. This witnesses nonemptiness and one half of normalization.

For the second half of normalization, any point with $\lambda +b<1$ would project down to make a point outside of $H_1$ and $H_2$, so it can't exist. Further, if condition 2 is fulfilled, you can make the a-measure $\left(\lambda \right(\mu \times {\mu }^{\prime }),b)$, with $\lambda +b=1$, and it projects down to $(\lambda \mu ,b)$ and $(\lambda {\mu }^{\prime },b)$ in $H_1$ and $H_2$ respectively, witnessing that it's present in the free product, and witnessing that the other half of normalization works out.

For closure and convexity and upper-completeness, the intersection of two closed convex upper-complete sets (the pullback of closed convex upper-complete sets along a continuous linear mapping that preserves the $b$ term) is closed-convex upper-complete, so that works out.

That leaves the compact-projection property, which is the only one where we may have issues. Remember, the necessary-and-sufficient conditions for compactness of a set of measures are: bounded $\lambda$ value, and, for all $ϵ$, there's a compact subset of $X\times Y$ where all the measure components of the points have all but $ϵ$ measure on said set.

Any point in the intersection of preimages with the measure component having a $\lambda$ value above the minimum of the two ${\lambda }_1^{\odot }$ and ${\lambda }_2^{\odot }$ values (maximum amount of measure present in points in $H_1$ and $H_2$ respectively) would project down to not be in $H_1$ or $H_2$, respectively, so those are ruled out, and we've got a bound on the amount of measure present.

As for an $ϵ$-almost-support for all the measure components in $H_1\ast H_2$, what you do is find a $C_{\frac{ϵ}{2}}^X$ compact $\frac{ϵ}{2}$-almost-support for $H_1$ and a $C_{\frac{ϵ}{2}}^Y$ compact $\frac{ϵ}{2}$-almost-support for $H_2$, and consider $C_{\frac{ϵ}{2}}^X\times C_{\frac{ϵ}{2}}^Y$

We claim this is a compact (product of compact sets) $ϵ$-almost-support for all measure components in $H_1\ast H_2$.

This is because any measure component in $H_1\ast H_2$ must project down to be a measure component of $H_1$ and a measure component of $H_2$. So, if there was more than $ϵ$ measure outside 

$C_{\frac{ϵ}{2}}^X\times C_{\frac{ϵ}{2}}^Y$

then it would project down to either $H_1$ or $H_2$ and have more than $\frac{ϵ}{2}$ measure outside of the two compact sets, which is impossible by the projection landing in $H_1$ (or $H_2$) and the two compact sets being $\frac{ϵ}{2}$-almost-supports for $H_1$ and $H_2$ respectively.

For any $ϵ$ we can make a compact subset of $X\times Y$ that accounts for all but $ϵ$ of the measure present of a measure in $H_1\ast H_2$, so it's got the compact-projection property, which is the last thing we need to certify it's an infradistribution.

Proposition 33: Having (homogenity or 1-Lipschitzness) present on both of the component infradistributions is a sufficient condition to guarantee the existence of the free product.

From Proposition 32, all we need to do is to verify properties 1 and 2, ie
1: There are points $(\lambda \mu ,0)\in H_1^{min}$ and $({\lambda }^{\prime }{\mu }^{\prime },0)\in H_2^{min}$ where $\lambda ={\lambda }^{\prime }$
2: There are points $(\lambda \mu ,b)\in H_1^{min}$ and $({\lambda }^{\prime }{\mu }^{\prime },b^{\prime })\in H_2^{min}$ where $\lambda ={\lambda }^{\prime }$, and $b=b^{\prime }$, and $\lambda +b=1$.

Given only that (without loss of generality) $H_1$ is 1-Lipschitz or homogenous, and same for $H_2$, we'll show properties 1 and 2.

The trick to this is showing that 1-Lipschitzness and homogenity both imply that an infradistribution contains some point with $\lambda =1$ and $b=0$, and if both infradistributions fulfill that, then conditions 1 and 2 are fulfilled for free, as can be verified with inspection.

So, let's show that 1-Lipschitzness and homogenity imply there's a point with $\lambda =1$ and $b=0$. For 1-Lipschitzness, $\lambda \le 1$, and since there must be a point with $b=0$ by normalization, and all points have $\lambda +b\ge 1$ by normalization, said point must have $\lambda =1$. As for homogenity, all minimal points have $b=0$, and normalization says there's a point with $\lambda +b=1$, getting that there's a point with $\lambda =1,b=0$. So, we're done.

Proposition 34: $p{r}_{X\ast }(h_1\ast h_2)\ge h_1$ and $p{r}_{Y\ast }(h_1\ast h_2)\ge h_2$, with equality when $h_1$ and $h_2$ are C-additive.