Equivalent form of the Univalence Axiom I'm reading the new HoTT book and I'm wondering about a potential equivalent form of the Univalence Axiom: $(A \simeq B) \simeq (A = B)$.
For simplicity, I'm tacitly working in a fixed universe. It is known that the univalence axiom implies function extensionality $$\mathsf{funext}:(f \sim g) \to (f = g)$$ where $f,g$ are from the dependent product type $\prod_{x:A} B(x)$ and $$(f \sim g) :\equiv {\textstyle\prod_{x:A} (f(x) = g(x))}$$ is the type of homotopies from $f$ to $g$. 
Assuming this, it seems that 
$$\mathsf{isequiv}(f) :\equiv \Big(\sum_{g:B\to A} (f \circ g \sim \mathsf{id}_B)\Big) \times \Big(\sum_{h:B\to A} (h \circ f \sim \mathsf{id}_A)\Big)$$
boils down to saying that $f$ is an isomorphism:
$$\mathsf{isisom}(f) :\equiv \sum_{g:B\to A} (f \circ g = \mathsf{id}_B)\times(g \circ f = \mathsf{id}_A).$$
Therefore, still assuming function extensionality, there is an equivalence between equivalence $$A \simeq B :\equiv \sum_{f:A \to B} \mathsf{isequiv}(f)$$ and isomorphism $$A \cong B :\equiv \sum_{f:A \to B} \mathsf{isisom}(f).$$
Assuming that the above is correct, the Univalence Axiom should be equivalent to the conjunction of functional extensionality with the weaker statement: $(A \cong B) \simeq (A = B)$. I like this conceptual splitting of the Univalence Axiom. Is the above correct or is there a flaw in my informal reasoning? Is it known that $(A \cong B) \simeq (A = B)$ is weaker than the Univalence Axiom?

As Mike pointed out in his answer, the correct definition of $A \cong B$ should be $$A \cong B :\equiv \sum_{f:A \to B} \mathsf{biinv}(f)$$ where $$\mathsf{biinv}(f) :\equiv \Big(\sum_{g:B\to A} (f \circ g = \mathsf{id}_B)\Big)\times\Big( \sum_{h:B\to A} (h \circ f = \mathsf{id}_A)\Big)$$
With this revised definition, it does appear to be the case that the Univalence Axiom splits into function extensionality and the formally weaker $(A \cong B) \simeq (A = B)$. The second question above should be corrected with this new formulation in mind: is $(A \cong B) \simeq (A = B)$ actually weaker than the Univalence Axiom?
 A: A nice thing about the usual formulation of UA in the form $(A = B)\simeq (A\simeq B)$ is that it implies $(A = B) = (A\simeq B)$, by UA one universe up.
A: Still not an answer to the second question, but I wanted to add something else that's missing: in fact the bare statement $(A=B)\simeq (A\simeq B)$ is not known to be a correct form of the univalence axiom.  The correct statement is that the canonical map $(A=B) \to (A\simeq B)$ is an equivalence.  The statement $(A=B)\simeq (A\simeq B)$ can be regarded as an abusive abbreviation of this, in the same way that mathematicians often write "$A\cong B$" to mean that some canonical map $A\to B$ is an isomorphism, but to be precise we have to be careful.
There is now a much weaker-looking known equivalent form of the univalence axiom, see here.  However, it does still use pointwise homotopies in the equivalences that are to be made into equalities.  I suspect that the corrected form "the canonical map $(A=B) \to (A\cong B)$" of your second question is indeed strictly weaker than univalence in the absence of funext, but I don't know.
A: You've made two changes in going from $\mathsf{isequiv}(f)$ to your "$\mathsf{isisom}(f)$": you replaced homotopy $\sim$ by equality $=$ (justified by function extensionality), and you condensed the left and right inverses $g$ and $h$ into a single inverse (not justified).  Thus, your "$\mathsf{isisom}(f)$" is not equivalent to $\mathsf{isequiv}(f)$ but rather to what we in the book called $\mathsf{qinv}(f)$, the type of quasi-inverses of $f$.  The two are logically equivalent, i.e. there are maps in both directions, but they are not equivalent types.  Thus, the univalence axiom does not even imply that $A=B$ is equivalent to your "$A\cong B$" — in fact, it contradicts it.  See sections 2.4, 4.1, and the rest of chapter 4.
If you made only the first change, replacing $\sim$ by $=$ but keeping $g$ and $h$ separate, then it would be true (by function extensionality) that $\mathsf{isequiv}(f)$ is equivalent to
$$
\Big(\sum_{g:B\to A}(f\circ g = \mathsf{id}_B)\Big) \times \Big(\sum_{h:B\to A}(h\circ f = \mathsf{id}_A)\Big),
$$
With this change, I think the rest of your reasoning is correct.
