Does $F$ is an isomorphism $\implies F^{-1}$ is an isomorphism? Does $F$ is an isomorphism $\implies F^{-1}$ is an isomorphism?
Let $\mathfrak{D, E} $ be some algebraic structures of the same 'kind' (say, groups, graphs, fields, vector spaces, etc.), and $F:\mathfrak D \to \mathfrak E$ is an isomorphism, is there any way to prove that $F^{-1}:\mathfrak E \to \mathfrak D$ is an isomorphism?
By isomorphism I mean a bijection that preserves whatever structure $\mathfrak D$ and $\mathfrak E$ have, clearly $F^{-1}$ is a bijection, so the important part is if it is a structure preserving map.
 A: The short answer is, "Yes, $F^{-1}$ will also be an isomorphism." This will be true for any sensible definition of isomorphism. Of course, in order to have a general proof, we will need a more precise definition of "algebraic structures" and "structure-preserving map". There are multiple ways to try to view this.
One way is "categories". These are collections of objects together with some "morphisms" between them. For example, in the category of groups, the morphisms are the group homomorphisms. The morphisms in a category are required to satisfy a few sensible axioms. In particular, it is possible to compose morphisms, and "identity" morphisms are required to exist. An isomorphism is a morphism $F: A \longrightarrow B$ that possesses an inverse morphism $F^{-1} : B \longrightarrow A$. By definition, this means that $F^{-1} \circ F$ and $F \circ F^{-1}$ are identity morphisms. From this it follows easily that $F^{-1}$ is also an isomorphism (with inverse $F$). Note that an isomorphism  in category theory is not defined as a bijective morphism. Indeed, "bijective" is not even a meaningful concept in all categories, and even in those where it is, "bijective morphism" is not necessarily the right definition of isomorphism.
However, the category definition of isomorphism may seem rather distant from the notion of "bijection that preserves whatever structure". Another way, closer to this, is using "varieties" from universal algebra. These generalize the notion of a group and ring to define a collection of sets that are required to satisfy certain axioms (such as the group axioms) for all members of the set.  However, these aren't as general as we'd like. (For instance, this doesn't cover fields, because in a field $0^{-1}$ doesn't exist.) We can generalize by considering structures in first order logic. However, this still doesn't cover topological spaces, an important example.
I am now going to present informally a rather general notion of structure. (This is based on Bourbaki's idea of a structure, which has not caught on and is not worth studying in detail unless you're rather curious. Most of what Bourbaki tried to do with structures should be done with categories instead. Here I just present enough of the idea in order to describe the notion of isomorphism.) The idea here is that given a set $A$, a structure on $A$ is a subset, satisfying the right rules (say, the group axioms), of a suitable set defined in terms of $A$. For example, a group on $A$ may be defined in terms of a subset of $A \times A \times A$, consisting of elements of the form $(g, h, g \cdot h)$, where $g \cdot h$ is the product of $g$ and $h$ in the group. And a topological space on a set $A$ is defined in terms of a subset of $\mathcal{P}(A)$ (the power set of $A$) satisfying the axioms for the open sets of a topological space.
Now it is possible to define "isomorphism". Consider the case of groups. Let $(A, \cdot_1)$ be a group. Then the set associated to $A$ is $S = \{(g, h, g \cdot_1 h) \mid g, h \in A\} \subseteq A \times A \times A$. And for another group $(B, \cdot_2)$, we have the set $T = \{(g, h, g \cdot_2 h) \mid g, h \in B\} \subseteq B \times B \times B$.  Given a function $F : A \longrightarrow B$, we obtain a function $F_0$ from $A \times A \times A$ to $B \times B \times B$, which maps $(g, h, k)$ to $(F(g), F(h), F(k))$. Then $F$ defined to be an isomorphism if it is a bijection and

$x \in S$ if and only if $F_0(x) \in T\quad\quad\qquad(*)$

holds. This latter condition is equivalent to asserting "$g \cdot_1 h = k$ if and only if $F(g) \cdot_2 F(h) = F(k)$".
This definition works for fairly general kinds of structures. The main thing is how to define $F_0$ from $F$. For example, if $F: A \longrightarrow B$ is a function between topological spaces, then $F_0: \mathcal{P}(A) \longrightarrow \mathcal{P}(B)$ is defined as $F_0(X) = \{F(x)\mid x \in X\}$ for $X \in \mathcal{P}(A)$ (the image of the set $X$ under $F$). If we define $S$ as the set of open sets in $A$ and $T$ as the set of open sets in $B$, an isomorphism is then defined as a bijection that satisfies the same condition $(*)$ above as for groups, and this definition is equivalent to "$F$ is a bijection, and $X$ is an open set in $A$ if and only if $F_0(X)$ is an open set in $B$".
In the general setup, there will always be a set $S$ that defines the structure of $A$ and a set $T$ that defines the structure of $B$. Then, an isomorphism $F$ will have an inverse that is also an isomorphism. This is because, by applying the function $F_0^{-1}$ to both sides of $(*)$, we see that $(*)$ implies

$F_0^{-1}(x) \in S$ if and only if $x \in T$

which is the condition for $F^{-1}$ to be an isomorphism.
A: yes. as you have stated, the map is bijective. The only thing left to prove is that it is a homomorphism, which shouldn't be to hard.
this is a duplicate
$F:G\to B$ is an isomorphism between groups implies $F^{-1}$ is an isomorphism
A: An isomorphism is just a relabeling with distinct labels, such that all the axioms hold after relabeling. So obviously the inverse is an isomorphism too, because the labeling can be undone.
For example, a group $(G,\circ,1_G)$ obeys:

$\forall x \in G\ \exists y \in G\ ( x \circ y = 1_G )$.

An isomorphism $f$ is essentially a relabeling of everything in $(G,\circ,1_G)$, meaning every member $x$ in $G$ becomes $f(x)$, and $\circ$ becomes $f(\circ)$, and $1_G$ becomes $f(1_G)$. So the axiom when relabeled becomes:

$\forall x \in f(G)\ \exists y \in f(G)\ ( x \operatorname{f(\circ)} y = f(1_G) )$.

If $f$ is an isomorphism from $(G,\circ,1_G)$ to $(H,\bullet,1_H)$, then we are effectively saying $f(G) = H$ and $f(\circ) = \bullet$ and $f(1_G) = 1_H$, so the axiom reduces to:

$\forall x \in H\ \exists y \in H\ ( x \bullet y = 1_H )$.

Similarly you can easily see that undoing the labeling gives the inverse isomorphism.
A: Suppose $f:D\to E$ is an isomorphism where $f(d_1)=e_1$ and $f(d_2)=e_2$.
Then we know $f(d_1d_2)=f(d_1)f(d_2)=e_1e_2$
Applying the inverse to both sides (which as you mentioned, must exist since $f$ is a bijection), we have
$(f^{-1}\circ f)(d_1d_2)=f^{-1}(e_1e_2)$
$d_1d_2=f^{-1}(e_1e_2)$
That is $f^{-1}(e_1e_2)=f^{-1}(e_1)f^{-1}(e_2)$ so $f^{-1}$ also preserves multiplication.
A: Isomorphic means two algebraic structures that are merely same and they have same property but different in naming or in some other perspective.
To prove two structures are isomorphic, we define a function in such a way to say that both structures are isomorphic, if they are isomorphic, naturally the domain and the codomain can be interchanged, thereby arriving your result. 
Since $F$ is isomorphism, it is one-one and onto, so naturally inverse $F^{-1}$ exists, since $F$ is one-one and onto, $F^{-1}$ is also one-one and onto. It remains to prove homomorphism part alone. $F^{-1}(a+b) = F^{-1}(a)+F^{-1}(b).$(For groups with addition operation)
