When do coherence spaces have pullbacks and pushouts? $\newcommand{\symp}{\Bumpeq}$

A coherence relation $\symp_X$ on a set $X$ is a reflexive and symmetric relation. A coherence space is a pair $(X, \symp_X)$, and a morphism $f : X \to Y$ between coherence spaces is a relation $f \subseteq X \times Y$ such that for all $(x,y) \in f$ and $(x&#39;,y&#39;) \in f$, 


if $x \symp_X x&#39;$ then $y \symp_Y y&#39;$, and 
if $x \symp_X x&#39;$ and $y = y&#39;$ then $x = x&#39;$.


The category of coherence spaces is both Cartesian and monoidal closed. I would like to know when pullbacks or pushouts exist for this category, and when some monoidal analogue of pullbacks or pushout exists (and how to define it, in case this notion makes sense).
I now see how to define equalizers for coherence spaces, which means pullbacks always exist (since products do). I don't know how to do this, actually....

Recall that composition is the usual relational composition, so if $f : A \to B$ and $g : B \to C$, then: 

$f ; g = \{(a,c) \in A \times C \;|\; \exists b \in B.\; (a,b) \in f \land (b,c) \in g\}$

(In this definition, the existential actually implies unique existence. Suppose that we have $b&#39; \in B$ such that $(a,b&#39;) \in f$ and $(b&#39;, c) \in g$. Since we know that $a \Bumpeq_A a$, this means that $b \Bumpeq_B b&#39;$. Then this means that we have $b \Bumpeq_B b&#39;$ and $(b,c) \in g$ and $(b&#39;,c) \in g$, so consequently $b = b&#39;$.)

We now construct equalizers. Suppose we have coherence spaces $A$ and $B$, and morphisms $f, g : A \to B$. Now define the equalizer $(E, e : E \to A)$ as follows. 


For the web, take 
$$E = \left\{ 
\begin{array}{l|c}
&amp; \forall b.\; (a,b) \in f \implies \exists a&#39; \Bumpeq_A a.\; (a&#39;,b) \in g \\
a \in A &amp; \land \\
&amp; \forall b.\; (a,b) \in g \implies \exists a&#39; \Bumpeq_A a.\; (a&#39;,b) \in f \\
\end{array}
\right\}
$$
This picks out the subset of tokens of $A$ on which either $f$ and $g$ agree (up to coherence -- I had this wrong in my first version), or are both undefined. 
Define the coherence relation on $\Bumpeq_E = \{ (a,a&#39;) \in \Bumpeq_A \;|\; a \in E \land a&#39; \in E\}$. This is just the restriction of the coherence relation on $A$ to the subset $E$. This will be reflexive and symmetric since $\Bumpeq_A$ is. 
The equalizer map $e$ is just the diagonal $e : E \to A = \{(a, a)\;|\;a \in E\}$. 


Since I messed up my first version of the proof, I'll give the universality property explicitly. Suppose we have any other object $X$ and morphism $m : X \to A$ such that $m;f = m;g$.

Now define $h : X \to E$ as $\{(x,a) \;|\; a \in E\}$. Obviously $h;i \subseteq m$, but to show the equality we need to show the converse $m \subseteq h;i$. 

So assume $(x,a) \in m$. We now need to show that $\forall b.\; (a,b) \in f \implies \exists a&#39; \Bumpeq_A a.\; (a&#39;,b) \in g$ and $\forall b.\; (a,b) \in g \implies \exists a&#39; \Bumpeq_A a.\; (a&#39;,b) \in f$. 

First, assume $b \in B$ and $(a,b) \in f$. So we know that $(x,a) \in m$ and $(a,b) \in f$, so $(x,b) \in m;f$. Therefore $(x,b) \in m;g$, and so there is an $a&#39; \in A$ such that $(x,a&#39;) \in m$ and $(a&#39;,b) \in g$. Since $x \Bumpeq x$, we know $a \Bumpeq a&#39;$, and so there is an $a&#39; \Bumpeq a$ such that $(a&#39;,b) \in g$. 

Symmetrically, assume $b \in B$ and $(a,b) \in g$. So we know that $(x,a) \in m$ and $(a,b) \in g$, so $(x,b) \in m;g$. Therefore $(x,b) \in m;f$, and so there is an $a&#39; \in A$ such that $(x,a&#39;) \in m$ and $(a&#39;,b) \in f$. Since $x \Bumpeq x$, we know $a \Bumpeq a&#39;$, and so there is an $a&#39; \Bumpeq a$ such that $(a&#39;,b) \in f$.