Gluing axiom of a TQFT In the book, Lectures on tensor categories and modular functors by Bakalov and Kirillov they construct a TQFT.

When they come to prove the gluing axiom, they just mention that "...This statement is of purely topological nature, and we omit its proof." And I don't see how to prove it.

Basically, they claim that if we cancel two coupons of a special link we get the same manifold  via surgery. (See the picture from page 87 of the book)

Could someone provide me a proof of the gluing axiom or give me a reference containing a complete proof? I read a Turaev's book but that proof is also not easy to understand.
Given topological spaces $\rm X, Y$ and $\rm Z$ with two continuous functions $f : \rm Z \to X$, $g : \rm Z \to Y$, it is always possible to build the push-out  $$ \rm X \coprod_{Z} Y$$ which is obtained by gluing $\rm X$ and $\rm Y$ along $\rm Z$.

It is defined as the quotient of the disjoint union by the equivalence relation $f(z) = g(z)$.

In your examples, all the gluings are just with one point i.e $\rm Z = \{\ast\}$.
Given topological spaces $\rm X, Y$ and $\rm Z$ with two continuous functions $f : \rm Z \to X$, $g : \rm Z \to Y$, it is always possible to build the push-out  $$ \rm X \coprod_{Z} Y$$ which is obtained by gluing $\rm X$ and $\rm Y$ along $\rm Z$.

It is defined as the quotient of the disjoint union by the equivalence relation $f(z) = g(z)$.

In your examples, all the gluings are just with one point i.e $\rm Z = \{\ast\}$.