The context for this question is my answer to this question on the inverse function theorem. I'll try to replicate as much of the necessary information as possible:
Let $E$ and $F$ be real Banach spaces, $f: E \to F$ be a $C^k$ map ($k \geq 1$), let $T:E \to F$ be a linear isomorphism of Banach spaces (so it is continuous with continuous inverse). Define a map $g: E \to E$ by \begin{equation} g = T^{-1} \circ f \end{equation} So $g$ is also $C^k$. Suppose now that there are open subsets $U$ and $W$ of $E$ such that the restricted map $g:U \to W$ is invertible with $C^k$ inverse $g^{-1}: W \to U$. Now, define $V = f[U]$. In my answer to the question, I implictly assumed that $V$ is an open subset of $F$.
I thought I understood why $V$ is open, but later when I tried to write down a proper proof, I got confused. My initial approach was to note that $f = T \circ g$, so it is "almost" a composition of homeomorphisms, and hence an open map. However this is flawed reasoning because $g$ isn't a homeomorphism. Rather, it is the associated restricted map $\gamma: U \to W$, defined by $\gamma(x) = g(x)$ which is a homeomorphism, where we use the subset topology (now I explicitly use a different letter to emphasise that the domain and target spaces are different). This difference in domain and target spaces is confusing me.
So my question is how can one prove that the set $V$ defined above is open relative to $F$? It would be very much appreciated if you could spell out any details/subtleties regarding the domain/codomain. Also, I think this is mostly a topological problem, so the linear structure of the spaces $E,F$ and the map $T$ are mostly irrelevant. As such if there is a proof using only the topological properties of the spaces/maps, I'd like to see that (but I'm not very adept at topology, so please keep the answers elementary).
Note that $g(U)=W$ which is hidden in the "restriction" condition and so
$$V=f(U)=T(g(U))=T(W)$$
And since $T$ is a homeomorphism then $V$ is open since $W$ is.