Is the product of a linear map and a tangent to 0 function a tangent to 0 function?

Question

Dealing with TVS I have encountered that in order to complete some proofs, it sufficed to show the exact same statement, so I was wonder if my proof of it is correct. First, some definitions.

Let $f:\mathbb{R}\rightarrow \mathbb{R}$, we will say that $f$ is $o(t)$ if $$\lim_{t\to 0}\frac{f(t)}{t}=0 $$ If $E,F$ are topological vector spaces, a map $\phi:E\rightarrow F$ is said to be tangent to $0$ is for every open neighborhood of $0\in F$, there exists an open neighborhood $V$ of $0\in E$ and some $f$ which is $o(t)$ such that $$\phi(tV)\subseteq f(t)W $$ Now, is the following statement true?

Let $A$ be a continuous linear map from $E$ onto $F$ and $\phi:G\rightarrow E$ be a tangent to $0$ function, then $A\circ \phi$ is a tangent to $0$ map.

Here is my attempt of a proof:

Proof: Let $W$ be an open neighborhood of $0\in F$, since $A$ is linear and continuous, then $A^{-1}(W)$ is an open neighborhood of $0\in E$. Let $V$ be an open neighborhood of $0\in G$ such that $$\phi(tV)\subseteq o(t)A^{-1}(W) $$ Apply $A$ to both sets and use linearity of $A$ to obtain $$(A\circ \phi)(tV)\subseteq o(t)W \quad \blacksquare$$ Is this reasoning correct? I cannot help thinking that there is something I am missing.

Answer 1

I think your proof is essentially correct, now that objects are clearly identified, I would write it as follows

Let $W$ be an open neighborhood of $0\in F$, since $A$ is continuous, $A^{-1}(W)$ is an open neighborhood of $0\in E$. Let $V$ be an open neighborhood of $0\in G$ and $f$ (of the $o(t)$ class) such that $$ \phi(tV)\subseteq f(t)A^{-1}(W) $$ Apply $A$ to both sets and use linearity of $A$ to obtain $$ (A\circ \phi)(tV)\subseteq f(t)W \quad \blacksquare $$

Hope it helps.