TITLE: Question on Isomorphism
QUESTION [0 upvotes]: When I think about isomorphism between 2 objects (let say set), i will think of that as you can find a way to change the label of the elements in one set so that it will equal to the other. More formally saying there exists a map between them which is linear (homomorphic), injective and surjective. 
I understand why it need to be injective and surjective, these conditions ensure that all the elements in the co-domain have there own image. But why this map need to be linear ??

REPLY [2 votes]: Structural mathematics divides mathematical objects into different types of objects. For example, we have sets, groups, vector spaces, topological spaces, metric spaces. For each type of object, we have the notion of a homomorphism or just morphism between two objects. For example:

A morphism of sets is a mapping.
A morphism of groups is a group homomorphism
A morphism of vector spaces is a linear map
A morphism of topological spaces is a continuous map
A morphism of metric spaces is an isometry.

We can compose morphisms, and every object has an identity morphism. A type of objects together with a (good) choice of morphisms is called a category. You can find the precise definition here, but you don't need this in order to understand the concept of an isomorphism (at least in the mentioned examples).
Definition. An isomorphism is an invertible morphism. (Wikipedia)
That is, a morphism $f : X \to Y$ is an isomorphism if there is a morphism $g : Y \to X$ such that the composite $X \xrightarrow{f} Y \xrightarrow{g} X$ is the identity of $X$ and the composite $Y \xrightarrow{g} X \xrightarrow{f} Y$ is the identity of $Y$. You should keep in mind the following picture:

For example, an isomorphism of sets is just a bijective map. For another example, an isomorphism of vector spaces $f : V \to W$ (over a fixed field) is by definition a linear map such that there is a linear map $g : W \to V$ such that $f \circ g = \mathrm{id}_W$ and $g \circ f = \mathrm{id}_V$. In particular, the underlying map of $f$ is bijective, and actually this suffices: If $f : V \to W$ is a linear map whose underlying map is invertible, then the inverse map is linear, so that $f$ is an isomorphism.
This is really just a "coincidence"$^1$ and doesn't hold for other categories: By definition an isomorphism of topological spaces (often called homeomorphism) is a continuous map which has an inverse continuous map. Not every bijective continuous map is an isomorphism. The simplest example is $(X,\wp(X)) \to (X,\{\emptyset,X\})$, $x \mapsto x$ where $X$ has at least two elements. Notice that this morphism is not the identity morphism (only its underlying map of sets is the identity).
There are similar counterexamples for the category of partial orders.
That being said, I hope that it's clear that isomorphisms are not just (what you still learn in many places) bijective structure-preserving maps. They are invertible morphisms. In fact, there are categories where we have no underlying sets at all so that "bijective" and "map" isn't well-defined at all. One of the most important examples is the homotopy category whose objects are topological spaces and whose morphisms are homotopy classes of continuous maps. In that category the isomorphisms are also known as homotopy equivalences. A homotopy equivalence has no canonical underlying map of spaces or sets, and even if we choose one it is usually not bijective. For example, the inclusion $\{0\} \to \mathbb{R}$ represents a homotopy equivalence.
$^1$Actually it is more than just a coincidence: Every monadic functor is conservative. This implies at once that bijective homomorphisms of groups, rings, vector spaces, Lie algebras, etc. are isomorphisms. It also implies that the forgetful functor from topological spaces to sets is not monadic. However, the forgetful functor from compact Hausdorff spaces to sets is monadic (see here), which in turn implies the useful and well-known fact that a continuous bijection between compact Hausdorff spaces is a homeomorphism.
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