Recently I created/removed some folders on Outlook desktop app.

When I synchronized my corporate account by manually refreshing under the folders screen, it didn't bring the new folders, nor removed the removed ones.

Is this a known bug?

I have a Motorola Milestone, and I remember in the past, at least the folder addition worked.


Question: On surjections, idempotence and axiom of choice
Answer: This is only a partial answer, but it's way too long to go in the comments.  I'll describe a permutation model (of ZF with atoms) that seems to give a counterexample $A$ for your statement.  It will be easy to see that $A$ maps onto $A\times2$, but I haven't proved that there's no bijection between $A$ and $A\times2$; all I can say is that it looks quite plausible.  If the example is correct, it will automatically transfer to ZF (without atoms) by the Jech-Sochor theorem.  So here's the model.

Start with a set $U$ of atoms in 1-1 correspondence with the product $2^\omega\times3$ of the Cantor set and the ordinal 3.  I'll write [x,i] for the atom that corresponds to the pair $(x,i)$, where $x\in2^\omega$ and $i\in3$.   Let $G$ be the group of those permutations of $U$ that leave the first component (in the Cantor set) of each atom unchanged.  So $G$ is the direct product of continuum-many copies of the symmetric group $S_3$.  To define the class of supports, let me first introduce the terminology "thread" for an $\omega$-sequence of atoms $[x_n,i_n]$ such that, for each $n$, $x_{n+1}$ is obtained from $x_n$ by deleting the first bit, i.e., $x_{n+1}(k)=x_n(k+1)$.  Let the supports be those subsets of $U$ that are covered by finitely many threads.  This choice of supports (along with the choice of $U$ and $G$) determines a permutation model $M$.

In $M$, let $A$ be the set of all the threads.  There is a surjection from $A$ to $A\times 2$, namely the map that sends any thread $T$ to the pair $(T',b)$ where $T'$ is obtained from $T$ by deleting its first term, say $[x_0,i_0]$, and $b=x_0(0)$.  (So $b$ is the one component in the real $x_0$ that is omitted when one forms $x_1$.)  This surjection is exactly 3 to 1, as it essentially just "forgets" the $i_0$ from $T$.  

I believe, but have not proved, that $M$ contains no injection from $A\times2$ into $A$.  The idea of the proof would be to assume there is such an injection, fix a finite number of threads that support it, and then look at what the injection could do with threads $T$ whose $x_0$ has no tail in common with anything from the support.

I further believe that the model would also work if I had used 2 instead of 3, but that the missing part of the proof will be easier with 3.  The point is that, with 2, each thread would support the "complementary" thread, consisting of the same reals $x_n$ but the other elements $i_n$ in 2. It would also support various "mixtures" of itself and the complementary thread.   With 3, a thread should support no threads other than itself and truncations of itself.

Going even further out on a limb, I also believe that the model would work if, instead of the whole Cantor space $2^\omega$, I used only the subspace consisting of the eventually periodic sequences of 0's and 1's and required threads also to be eventually periodic.  In this model, $A$ should be Dedekind-finite, thus answering the last part of your question.