Let $V , W_1$, and $W_2$ be finite-dimensional vector spaces, and let $W = W_1 \times W_2$. Given an open set $U \subseteq V$ and functions $f_1 :U \rightarrow W_1$ and $f_2 :U \rightarrow W_2$, the rule $f(x)=(f_1(x),f_2(x))$ defines a mapping from $U$ into $W$. Let $a \in U$.
Prove that $f$ is differentiable at $a$ if and only if $f_1$ and $f_2$ are differentiable at $a$.
$(\Rightarrow)$ Suppose $f$ is differentiable at $a$. Let $T:V\rightarrow W$ be a linear mapping. Then there exists $T_1:V\rightarrow W_1$ such that $f_1=T_1\circ f_1$. Then $f_1=T_1\circ f_1\implies$
$(Df_1)(a)=D(T_1\circ f_1)(a) = (DT_1\circ Df_1)(a) = (T_1\circ Df)(a)$. Therefore $Df_1$ exits. The same argument is made for $f_2$.
$(\Leftarrow)$ Suppose $f_1$ and $f_2$ are differentiable at $a$. Let $T_1:V\rightarrow W_1$ and $T_2:V\rightarrow W_2$ be a linear mappings. Then there exists $T:V\rightarrow W$ such that $T(v)=(T_1(v), T_2(v))$ that is linear.
Is the forward part of my proof correct? I am trying to figure out the converse, but I want to be sure the first part is correct before spending too much time on it.