Why does this diagram not commute? When I look at the following commutative Diagram involving two vector spaces and the linear map $S$, why does it not seem to commute? Should it even commute?

For example if $\text{dim} V = 2$ und $\text{dim} W = 1$, then the rank of $S^* = 1$ but rank of the map to its dual space could be $2$.
This somehow confuses me, why does it behave that way?
 A: I am understanding your question as asking about vector spaces $V,W$, their duals $V^*$ and $W^*,$ a linear map $S$ and its dual $S^*$ and given isomorphisms from $V$ to $V^*$ and $W$ to $W^*.$ In general, this diagram is not expected to commute. This noncommutativity is related to the fact that it is generally impossible to choose a canonical isomorphism from a vector space to its dual. That is to say that isomorphisms from $V$ to $V^*$ and $W^*$ cannot be chosen in a unique way and will likely have nothing to do with each other. The situation is different if we are considering canonical isomorphisms from two vector spaces to their double dual. See, for example, Motivation to understand double dual space.
A: Yes, it is fairly surprising that no broad version of such a family of diagrams can commute.
Even if we restrict out attention to finite-dimensional Hilbert spaces over $\mathbb R$, and use the Riesz-Fischer isomorphism of $V$ to $V^*$, the analogous diagram does not commute unless $S$ is an isometry to its image. This is already clear with two-dimensional real Hilbert spaces.
For emphasis: it did surprise me considerably some years ago when I came to understand that the Riesz-Fischer isomorphism is generally "not functorial" ... in the sense that the expected diagram fails to commute. For me, this arose in looking at things like $H^1\to L^2\to H^{-1}$ where $H^i$ are Sobolev spaces, and $H^{-1}$ is dual to $H^1$... Details about this would probably be too tangential to the basic question here.
