What is the reason that isomorphisms between vector spaces are invertible? I ask this soft-question because I am confused about the intuition of the mechanics of proving something is an isomorphism:
My professor gives the following steps to check for isomorphism:

*

*The function is one-to-one

*The function is onto

*The function preserves addition

*The function preserves multiplication

However, I looked up another video to further clarify the intuition behind isomorphism, and the video states: "Suppose V and W are vector spaces over the same field. We say that $V$ and $W$ are isomorphic if there exists an invertible linear transformation  $T: V\to W$"
I think I understand why the preservation of addition and multiplication fits into the "invertibility" as they are commutative. But I don't know how the "1-to-1 and onto (i.e., bijection)" contribute to invertibility as defined by the video?
Thanks!
 A: We say that two vector spaces $V,W$ are isomorphic if there is a function $T : V \rightarrow W$ which is an isomorphism. Your question seems to be about different definitions of what it means to be an isomorphism, so I'll compare these definitions.
First, let me rephrase the definition your professor gave you. A function which is both one-to-one (injective) and onto (surjective) is called bijective. Moreover, a linear function is precisely one which preserves addition and (scalar) multiplication. So, your conditions for $T$ to be an isomorphism are equivalent to the two following conditions:

*

*(A) $T$ is bijective;

*(B) $T$ is linear.

Note that conditions 1. and 2. have combined to form condition (A), and conditions 3. and 4. have combined to form condition (B).
The definition which you found online, however, says that $T$ is an isomorphism if it is an invertible linear transformation. What does it mean for a linear transformation $T : V \rightarrow W$ to be invertible? It means that there exists another linear map $S : W \rightarrow V$ such that $S \circ T = Id_V$ and $T \circ S = Id_W$. How do we convince ourselves that these two definitions are equivalent?
First, let's forget about linearity for a moment, and just consider any old function $f : A \rightarrow B$ between two sets. It turns out that $f$ is bijective if and only if there exists a function $g : B \rightarrow A$ such that $g \circ f = Id_A$ and $f \circ g = Id_B$ (this function $g$ is called the inverse of $f$). This is the first important fact here, and it's something that you should try to prove yourself; if you get stuck, you can probably find another question on this site that will help.
Now, let's try to prove the equivalence of our two definitions.

*

*On one hand, if we assume that $T$ is an invertible linear transformation, then of course $T$ is linear, and existence of a (linear) inverse tells us that $T$ is bijective, because of the fact I noted in the previous paragraph. So, conditions (A) and (B) are satisfied.

*On the other hand, if we assume conditions (A) and (B), then $T$ is linear, and the fact that it's bijective lets us conclude that it has an inverse $S$. However, we're not quite done yet, because the definition of "invertibility" that I gave requires that $S$ is in fact a linear inverse. But, here's the second important fact: if a linear map $T$ is bijective, then its inverse is necessarily linear as well. (I'd recommend trying to prove this yourself too, but you can also check out the answers to this question.) So, we can conclude that $T$ really is invertible.

So, the two definitions you brought up are equivalent. I agree that the four-point definition you gave is generally more useful for proving directly that a given map is an isomorphism, but the invertibility condition can be more useful in certain situations, so it's good to know both.
