Following a lecture https://www.youtube.com/watch?v=hZHnUcy7rNc&list=PL3GPZrYvP8TjHuKUzrtUPiu_dwn8kYcqS&index=1&ab_channel=NDGeoTop on index and K-theory, the following definition is given at 58:00 for the relative K-theory of a pair $X\supseteq A$ with $X$ a compact manifold and $A$ open, as: \begin{equation} K(X,A)=\big\{ (E^+ \xrightarrow{\alpha} E^-) | \alpha \small|_A \text{is an isomorphism}\big\}\big/\sim \end{equation} where $\sim $ is an equivalence relation identifying (i) isomorphic $\alpha$'s and (ii) triples $(E^+ \oplus F \xrightarrow{\alpha \circ id_F} E^- \oplus F)$ for any $F$ over $X$. In other words it is a triple (instead of a couple $(E^+, E^-)$ of the original K-theory of $X$) such that the pairs $(E^+,E^-)$ are trivial by the existence of a bundle map on the base space $X$ that is a bundle isomorphism on the subspace $A$.
I would like to know if this restricts the dimension of $E^+$ and $E^-$ to be the same on connected components of $X$ to which $A$ belongs. I have in mind that in the $K$-theory of $X$, the pair of bundles $(E^+,E^-)$ can be of different dimensions. But the fact that $\alpha$ is an isomorphism on A, means that $E^+$ and $E^-$ should have the same dimension over $A$, and since they are vector bundles, the dimension is an invariant on connected components (?).
It is not clear to me how this definition relates to the definition $K(X,A)=\tilde{K} (X/A)$. Please, bring some elements. In fact I am not sure to understand correctly the definition of the reduced K-theory of X with the inclusion map.
PS: I am a physicist.