$T: R^3->P_{3}$ T([a b c]) = $at + bt^2 + ct^3$
I am not sure if this is an isomorphism or not. The dimensions are not equivalent to each other. $dim(P_{3}) = 4$, but there are only 3 independent "vectors" within $at + bt^2 + ct^3$. So, I would say that T is an isomorphism.
How about for this one?
$T: R^4->P_{2}$ T([a b 0 c]) = $at + bt^2 + ct^3$ . I would also assume it is an isomorphism, for the same reasons.
Another function that I am unsure of is:
$T: C->R^2$: T([a b*i]) = $[a+b, a+b]$
I would say that the span of R^2 in this case would be the span of [1,1], hence not an isomorphism.
So I guess my main question is, T is an isomorphism if and only if dim(V) = dim(W) where dim(V) is the number of free variables or the dimension of the space where T is defined at?
To determine isomorphism between sets, you need to find a mapping from one set to the other that is both injective and surjective. For the first case, you cannot find an isomorphism because the dimensions are not equal, as you have stated. No such mapping can be surjective (why?).
In the second case, again the dimensions are not equal. No mapping can be injective (again, why?).
The last case is not quite correctly stated. I assume this should be $T:\mathbb C\rightarrow \mathbb R^2$ given by $T(a+bi) = (a+b, a+b)$. You are correct to say that the image of $T$ is the span of $(1, 1)$, and this is clearly not the same as $\mathbb R^2$. Hence, the mapping is not surjective.
In response to the main question, not quite. $T: V\rightarrow W$ is an isomorphism iff it is injective and surjective. A mapping is injective iff the dimension of the kernel is 0, and a mapping is surjective iff the dimension of the image equals the dimension of the codomain, $W$. If $\dim V = \dim W$, then by the rank-nullity theorem, knowing either injectivity or surjectivity is enough to ascertain the other. But knowledge of one is still required.
For example, consider $T:\mathbb R^3 \rightarrow \mathbb R^3$ given by $T(x, y, z) = (0, 0, 0)$. The dimensions of $\dim V$ and $\dim W$ are equal, but the dimension of Ker$(T) = 3$, hence the dimension of Im$(T) = 0$. Thus, $T$ is not an isomorphism.