In Hirsch's Differential Topology Chapter 8 Theorem 2.3, it says :
Let $f, g:\partial Q\approx \partial P$ be isotopic diffeomorphisms. Then $P\cup_f Q\approx P\cup_g Q$. Here $\approx$ means diffeomorphic.
Now the question I have is the following. Let's say I have a closed 4-ball and its boundary $S^3$. Consider a knot $k\subseteq S^3$, and a tubular neighborhood of it, $N(k)\subseteq S^3$. Then I can glue a 2-handle $D^2\times D^2$ along $S^1\times D^2$, if I am given a diffeomorphism $h:S^1\times D^2\to N(k)$.
By Hirsch's, isotopic $h$'s give diffeomorphic manifolds.
Now Hirsch also says that any two tubular neighborhoods of $k$ in $S^3$ are isotopic (Chapter 4, Theorem 5.3).
The problem then is if we have two tubular neighborhoods $f_0:S^1\times D^2\to N(k)$ and $f_1:S^1\times D^2\to N(k)$, this seems to imply that $f_0$ and $f_1$ are isotopic. This would effectively mean that the choice of framing for the knot doesn't change the diffeomorphism type of the surgered manifold.
How is this argument wrong? I understand this is about discerning between different notions of the statement "tubular neighborhoods are unique up to isotopy", which is what I haven't been able to do.
The definition of an "isotopy of tubular neighborhoods" (bottom of page 111) is not what you might be expecting: yes it's an isotopy, but it only needs to carry one tubular neighborhood to another by a vector bundle isomorphism. This means that if you have a knot $K\subset S^3$ with two framings $h_1,h_2:S^1\times D^2\to N(K)$, then while there is an isotopy of tubular neighborhoods between both framings' tubular neighborhoods, there's certainly no reason to expect $h_2^{-1}\circ h_1:S^1\times D^2\to S^1\times D^2$ to be isotopic to the identity. (This composition is describing the difference between gluing along $h_1$ and gluing along $h_2$.)
Another way to say it is that the maps $h_1$ and $h_2$ are not necessarily isotopic, even if the tubular neighborhoods they describe are isotopic as tubular neighborhoods.