\newcommand{\ann}{\operatorname{ann}}
\newcommand{\rad}{\operatorname{rad}}
\newcommand{\gen}{\operatorname{span}}
\newcommand{\im}{\operatorname{im}}
\noindent{\bf\red Proof of Theorem 1.}

Uniqueness: If $A$ and $B$ are 2 pushforwards, then $\sigma_W(U+A) = \sigma_W(U+B)$ for all PQs $U$ on $W$.

Thus $\calD_A = \calD_B$, because otherwise if $w \in \calD_A\setminus \calD_B$, by taking $U(w) = 1$ on $\calD_U = \gen\{w\}$, we get $\sigma_W(U+A) = 1 \neq 0 = \sigma_W(U+B)$. Furthermore, $A$ and $B$ must agree where they are both defined, because by taking $U(w) = \frac{-A(w)-B(w)}{2}$ on $\calD_U = \gen\{w\}$ we get $(U+A)(w) = \frac{A(w)-B(w)}{2} = -(U+B)(w)$, so we must have $A(w) = B(w)$ to satisfy $\sigma_W(U+A) = \sigma_W(U+B)$. \\


Existence: Define $\phi_*Q$ by $\calD_{\phi_*Q}  =\phi (\operatorname{ann}_Q(\ker\phi))$ and $\phi_*Q(w) = Q(v)$ where $v \in \ann_Q(\ker\phi)$. Note that $\phi_*Q$ is well-defined. 

First consider when $U=0$ on all of $W$. Let $K$ be a maximal non-degenerate subspace of $\ker\phi$. Then $Q = Q\vert_K \oplus Q\vert_{\ann_Q(K)}$, and we can write $\ann_Q(K) = R \oplus A \oplus B$ where $R = \rad_Q(\ker\phi)$ and $A, B$ are chosen so that $A \subseteq \ann_Q(R)$ and $B \subseteq \ann_Q(K) \setminus \ann_Q(R)$. Since $Q: R \to B^*$ is surjective, for any $v \in \calD_Q$ there is some $r_v  \in R$ such that $Q(r_v, B) = Q(v, B)$. If we choose the $r_v$ so that $r_{v_1}+r_{v_2} = r_{v_1+v_2}$, then we can replace $A$ by $A' = \{a-r_a : a \in A\}$ and $B$ by $B' = \{b-\frac{1}{2}r_b : b \in B\}$ to get $Q = Q\vert_K \oplus Q \vert_{R \oplus B'} \oplus Q\vert_{A'}$. Then notice that\begin{itemize} \item $\sigma_V(Q\vert_K) = \sigma_{\ker \phi}(Q\vert_{\ker\phi})$ 
\item  $\sigma_V(Q \vert_{R \oplus B'}) = 0$
\item $\sigma_V(Q\vert_{A'}) = \sigma_W(\phi_*Q)$ \end{itemize} 
so we get $\sigma_V(Q) = \sigma_{\ker \phi}(Q\vert_{\ker\phi}) + \sigma_W(\phi_*Q)$. 

Now for an arbitrary $U$, note that $(Q + \phi^*U)\vert_{\ker \phi} = Q\vert_{\ker \phi}$ and $\phi_*(Q + \phi^*U) = \phi_*Q + U$ so we can replace $Q$ in the $U=0$ case by $Q + \phi^*U$ to get the general case. 


\noindent{\bf\red Proof of Theorem 2.}

It's clear that pullback is functorial and that pushforward by the identity is the identity. To show $(\phi\psi)_*=\phi_*\psi_*$, use theorem 1 repeatedly to get \begin{align*} &\sigma((\phi\psi)_*Q + U)\\ =&\sigma(Q + (\phi\psi)^*U) \\=&\sigma(Q + \psi^*\phi^*U)-\sigma(Q\vert_{\ker \phi\psi}) \\
=& \sigma(\psi_*Q + \phi^*U)+\sigma(Q\vert_{\ker \psi})-\sigma(Q\vert_{\ker \phi\psi}) \\
=& \sigma(\phi_*\psi_*Q+U)+\sigma(Q\vert_{\ker \psi})+\sigma(\psi_*Q\vert_{\ker \phi})-\sigma(Q\vert_{\ker \phi\psi}) \\
=&\sigma(\phi_*\psi_*Q+U)\end{align*}
for any $U$, where the last step uses theorem 1 on $Q\vert_{\phi\psi}$ with the map $\psi: \ker\phi\psi \to \ker\phi$. 

\parpic[r]{$\xymatrix@C=3mm@R=3mm{
  \bullet \ar[r]^\alpha \ar[d]_\gamma & \bullet \ar[d]^\beta \\
  \bullet \ar[r]_\delta \ar@{.>}[ru] & \bullet
}$}
To show $\alpha_*\gamma^* = \beta^*\delta_*$, first note that $\beta^*\beta_*$ is the identity on any PQ since $\beta$ is injective, so $$\alpha_*\gamma^*Q =  \beta^*(\beta\alpha)_*\gamma^*Q =\beta^*(\delta\gamma)_*\gamma^*Q= \beta^*\delta_*\gamma_*\gamma^*Q\hspace*{1cm}$$

As $\beta^*\delta_*\gamma_*\gamma^*Q$ and $\beta^*\delta_*Q$ have the same values where they are both defined, it remains to show that they have the same domain. Since $\alpha$ is surjective and $\gamma$ is surjective onto $\ker(\delta)$, we see that $$\beta^{-1}\delta(A) = \beta^{-1}\delta(A \cap \im\gamma)$$ for any subspace $A$. By taking $A = \ann_Q(\ker\delta)$, the two sides of the equality become the domains of  $\beta^*\delta_*Q$ and $\beta^*\delta_*\gamma_*\gamma^*Q$.

 %first note that if $\phi$ is injective, then $\phi_*\phi^*U = \phi_*\phi^*U = U$ for any $U$. Then use theorem 1 repeatedly and injectivity of $\beta$ and $\gamma$ to get
%\begin{align*}
%&\sigma(\alpha_*\gamma^*Q + U) \\
%=&\sigma(\gamma^*Q + \alpha^*U) - \sigma(\gamma^*Q\vert_{\ker \alpha})\%\
%=&\sigma(\gamma^*Q + (\beta\alpha)^*\beta_*U) - \sigma(\gamma^*Q%\vert_{\ker \alpha})\\
%=&\sigma(\gamma^*Q + (\delta\gamma)^*\beta_*U) - \sigma(\gamma^*Q%\vert_{\ker \alpha})\\
%=&\sigma((\delta\gamma)_*\gamma^*Q + \beta_*U) +\sigma(\gamma^*Q%\vert_{\ker\delta\gamma})- \sigma(\gamma^*Q\vert_{\ker \alpha})\\
%=&\sigma(\delta_*Q + \beta_*U) +\sigma(\gamma^*Q\vert_{\ker\delta%%vv\gamma})- \sigma(\gamma^*Q\vert_{\ker \alpha})\\
%%=&\sigma(\beta^*\delta_*Q +U)-\sigma(U\vert_{\ker\beta}) +%\sigma(\gamma^*Q\vert_{\ker\delta\gamma})- \sigma(\gamma^*Q\vert_{\ker %\alpha})\\
%=&\sigma(\beta^*\delta_*Q +U)
%\end{align*}