Does same cardinality imply a bijection?

"Same cardinality" is defined as meaning there is a bijection.

In your vector space example, you were requiring the bijection to be linear. If there is a linear bijection, the dimension is the same. There is a bijection between $\mathbb R^4$ and $\mathbb R^3$, but no such bijection is linear, or even continuous. (Space-filling curves, which are continuous functions from a space of lower dimension to a space of higher dimension, are not bijections since they are in no instance one-to-one.) If there is a bijection, then the cardinality is the same. And conversely.

I asked myself something similar: We have two sets $A$ and $B$ and we have two injections $f,g$ i.e. $$ f:A\rightarrow B $$ and $$ g:B\rightarrow A $$ is this enough to conclude that both sets have the same cardinality? In fact I found this theorem (Cantor-Bernstein-Schröder) very illuminating and the answer is then, yes, it is enough.

If we have the two injections $f,g$ we can find a bijection $h$ between both sets which therefore means they have the same cardinality.

So if you find two injections in your example, then you got your answer. Unfortunately it is quite hard to find an injection $\mathbb{R}^4\to \mathbb{R}^3$. The example given below does not work.

[EDIT as pointed out in the comments,$g$ is actually not an injection, it is only an injection from $S^1\times S^1\to \mathbb{R}^3 $ check here, page 2 bottom] $$ g:\mathbb{R}^4\to \mathbb{R}^3, g(x_1,x_2,x_3,x_4)=((x_1(2 + x_3),x_2(2 +x_3), x_4) $$ and $$ t:\mathbb{R}^3\to \mathbb{R}^4,t(x_1,x_2,x_3)=(x_1,x_2,x_3,0) $$

Use the positional expansion–based mapping proposed by @StevenGubkin in this comment $$(0.a_1 a_2 a_3 \ldots,\, 0.b_1 b_2 b_3 \ldots) \mapsto 0.a_1 b_1 a_2 b_2 a_3 b_3 \ldots$$ with a quite simple fix:

1. do not use expansions with infinitely repeating $9$ (if you use decimal system);
2. do interleave single digits only if they are not $9$, but for nines take all consecutive nines as an interleave unit together with the following single non–$9$ digit.

This makes the mapping a bijection $\Bbb R^2 \supset [0,1)^2\leftrightarrow [0,1)\subset\Bbb R$. Extension to all reals is quite straightforward, for example by $\tan$ and $\arctan$.