Derivative as sum of partial derivatives: Dieudonné's proof The following is proposition (8.9.1) from Jean Dieudonné's "Foundations of Modern Analysis", 1960, which is on the net:
https://archive.org/details/FoundationsOfModernAnalysis_578, pp. 167-168. 


Concerning part b) of Dieudonné's proof, where it is shown that Df exists and is continuous on A, if $ D_1f $ and $ D_2f $ exist and are continuous on A, I have three questions, the third one being by far the severest. 
To the first question: My understanding is that the 5th line of part b) follows from the fact that $ D_1f $ exists. Then Dieudonné makes use of his version 8.6.2 of the mean value theorem and of the continuity of $ D_2f $ to show that

My question is: Why does this last line not follow directly from the fact that $ D_2f $ exists, in analogy to what we did in the 5th line for $ D_1f $? $ D_2f $ not only exists at the point $ (a_1,a_2) $, but on all of A, so also at the point $ (a_1+t_1,a_2) $, assuming that $ t_1 $ is small enough. Why do we have to take the detour with the MVT? 
PS: This question I consider to be answered already by the two comments by Ted Shifrin following this post.

The second question is: Why do we need $ 4 \epsilon $ in the 3rd last line? My understanding is that we can decompose this statement into 3 parts (take the 2nd and 3rd line of part b as a hint to this decomposition), namely the three parts which Dieudonné describes previously, and that $ 3 \epsilon $ would be sufficient. Is this a typo in the book?

The third question concerns the last two lines of the proof of (8.9.1), where the continuity of $Df$ is derived from the continuity of $D_1f$ and $D_2f$. To me the given equation 
$Df = D_1f \circ i_1 + D_2f \circ i_2 \qquad $    (1)
as it stands makes no sense. Its three terms are maps of different type which cannot be combined in this way. I suppose that Dieudonné assumes the same argument $(a_1,a_2)$ to be added to each term, because a similar short-hand equation is given in the following proposition (8.9.2). However, the resulting equation
$Df(a_1,a_2) = D_1f(a_1,a_2) \circ i_1 + D_2f(a_1,a_2) \circ i_2 \qquad $    (1a)
still does not work in my opinion and does not properly represent (8.9.1.1) as he claims. Is there another typo here? In my proposed answer to this question I want to show that we need a slightly different equation. In order to demonstrate that it is correct, I will also use it to proof the equation of proposition (8.9.2). 
Is all this correct? 

Thanks for help!
 A: Proposed answer to the third question (please see the comment to this answer):
In order to explain why we need an equation different from the one Dieudonné gives, we have to go back to the end of part a) of the proof of (8.9.1), where the continuity of $D_1f$ and $D_2f$ is derived from the continuity of $Df$. It took me some time to (hopefully) understand Dieudonné's reasoning here. It is based on the equation 
$D_1f(a_1,a_2)=Df(a_1,a_2) \circ i_1. \qquad $ (2) 
He constructs the function $ \alpha: (v,u) \rightarrow v \circ u $ as described to be of type $ \mathcal{L}(E_1 \times E_2,F) \times \mathcal{L}(E_1, E_1 \times E_2) \rightarrow \mathcal{L}(E_1,F) $. Then what I think he does is to represent $D_1f$ as the composition
$D_1f = \alpha(Df(a_1,a_2),i_1) \circ Df. \qquad $ (3)
With this notation (sorry, I can't think of a better one) I mean that the argument $(a_1,a_2)$ is applied to Df, and then to the result $Df(a_1,a_2)$ we apply $i_1$. Then on the right hand side of (3) we obtain a map $ A \subseteq (E_1 \times E_2) \rightarrow \mathcal{L}(E_1,F), \; \; (a_1,a_2) \mapsto D_1f(a_1,a_2)$. Now the continuity of $D_1f$ follows from the continuity of Df and the continuity of $\alpha$. I'm not sure whether we need the continuity of $i_1$ here, too, but we have it anyways.
Let's go back to our original problem. In equation (1) we have three maps of different type
$Df: E_1 \times E_2 \rightarrow \mathcal{L}(E_1 \times E_2,F) \\
D_1f \circ i_1: E_1 \rightarrow \mathcal{L}(E_1,F) \\
D_2f \circ i_2: E_2 \rightarrow \mathcal{L}(E_2,F)$
which don't fit together. Equation (1) is supposed to follow from (8.9.1.1.) which I think is not true, either. However, we can take the following equation instead, which in my opinion is a true rewording of (8.9.1.1):
$ Df(a_1,a_2) = D_1f(a_1,a_2) \circ pr_1 + D_2f(a_1,a_2) \circ pr_2 \qquad $ (4)
where $pr_1$ and $pr_2$ denote the projections $(t_1,t_2) \mapsto t_1$ and $(t_1,t_2) \mapsto t_2$. So basically what we do is, we use the projections $pr_k$ instead of the natural embeddings $i_k$. In my opinion here we simply have another typo in the book. Now we can represent $Df$ as follows
$ Df= \beta(D_1f(a_1,a_2),pr_1) \circ D_1f + \gamma(D_2f(a_1,a_2),pr_2) \circ D_2f \qquad $ (5)
with $ \beta: (v,u) \rightarrow v \circ u $ 
being a map $ \mathcal{L}(E_1,F) \times \mathcal{L}(E_1 \times E_2,E_1) \rightarrow \mathcal{L}(E_1 \times E_2,F) $
and $ \gamma: (v,u) \rightarrow v \circ u $ 
being a map $ \mathcal{L}(E_2,F) \times \mathcal{L}(E_1 \times E_2,E_2) \rightarrow \mathcal{L}(E_1 \times E_2,F) $.
The right hand side of (5) is the sum of two maps of the type $ A \subseteq (E_1 \times E_2) \rightarrow \mathcal{L}(E_1 \times E_2,F) $, and the continuity of $Df$ follows by a reasoning analogous to the one used in part a) of the proof.

We can make use of equation (4) to easily prove the equation of proposition (8.9.2), too. As Dieudonné points out, (8.9.2) follows from (8.9.1) with the help of the chain rule. So we call $h:f \circ (g_1,\dots,g_n)$ and write with $b \in B$
$ Dh(b)=D \big( f \circ (g_1,\dots,g_n) \big) (b) = Df \big( (g_1,\dots,g_n) (b) \big) \circ D (g_1,\dots,g_n) (b) $
$ = \left( \sum_{k=1}^{n} D_kf \big( (g_1,\dots,g_n) (b) \big) \circ pr_k \right) \circ \big( Dg_1(b),\dots,Dg_n(b) \big) $
$ = \sum_{k=1}^{n} \big( (D_kf) \circ (g_1,\dots,g_n) (b) \big) \circ Dg_k(b).  $
The proof that Dh is continuous is a bit more tricky than in (8.9.1), I haven't figured that out completely yet.
