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Let $f : A \rightarrow B$ is the statement $f$ is a bijection if and only if for each partition $\left\{A_j \right\}_{j \in J}$ then $\left\{ f(A_j) \right\}_{j\in J}$ is a partition for $B$ true?

I think it is because... if $f$ is a bijection then $\forall x,y \in A, x \neq y \Rightarrow f(x) \neq f(y)$, I can then conclude then $\forall i,j, i \neq j, f(A_i) \neq f(A_j)$ and then $B = f\left(\bigcup_{j\in J} A_j \right) = \bigcup_{j\in J} f(A_j)$, here $\left\{ A_j \right\}_{j \in J}$ is a partition for $A$. Conversely instead I can consider the partition formed by sets of the form

$$ A_x = \left\{ x \right\} $$

where $x \in A$. So $f(A_x) = \left\{f(x)\right\}$ and $\bigcup_{x \in A} f(A_x) = B$ with $f(A_x) \cap f(A_y) = \emptyset$ when $x \neq y, \Rightarrow f(x) \neq f(y)$. Can the statement be restricted to something like $f$ is a bijection if and only if there is a partition of $A$ such that $\left\{ f(A_j) \right\}_{j \in J}$ is a partition for $B$, with $f|_{A_j}$ bijection itself? I think this is also true since:

The case $f$ is a bijection then there's a partition $\left\{ A_j \right\}_{j \in J}$ such that $f |_{A_j}$ is a bijection is easy to prove by just considering the partition which elements are $A_x = \left\{ x \right\}$, $x \in A$. Conversely let's consider a partition $\left\{ A_j \right\}_{j \in J}$ with such features, let's pick $x,y \in A$ such that $x \neq y$ then there are two indices $j_x,j_y \in J$ such that $x \in A_{j_x}$ and $y \in A_{j_y}$. If $j_x = j_y$ since the restriction to $A_{j_x}$ is a bijection then $f(x) \neq f(y)$, otherwise since by hypothesis $\left\{ f(A_j) \right\}_{j \in J}$ is a partition for $B$ then $f(A_{j_x}) \cap f(A_{j_y}) = \emptyset$ and then $f(x) \neq f(y)$ so $f$ is a bijection.

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    $A_x = \{\{x\} \;:\; x \in A\}$. – 2017-01-03
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    With $A_x$ I meant an element of a partition (which is a subset of elements of $A$). – 2017-01-03
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    If $x\in A$ then $x$ isn't necessarily a subset of $A$. $\{x\}$ is a subset of $A$ and $\{\{x\}\}_{x\in A}$ is a partition of $A$. – 2017-01-03
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    In any case, *your* $A_x = A$ for all $x$ and using $x$ free and bounded is horribly confusing. – 2017-01-03
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    You're right, it is confusing. What I meant to denote was something like $A_x = \left\{ x \right\}$ where $x \in A$. So for each $x \in A$ we have that $A_x$ is the subset of $A$ containing only $x$, is this ok? – 2017-01-03
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    I changed the notation a bit, hoping that $A_x$ is not confusing anymore. – 2017-01-03
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    Now makes sense. – 2017-01-03
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    So is the proof correct? Could have it been simpler than that? sometimes I think I prove things in a too twisted way... – 2017-01-04

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Only two little problems. In the first case: $$\forall i,j, i\ne j, f(A_i)\ne f(A_j)$$ isn't enough. $f(A_i)\cap f(A_j) = \emptyset$ is required (but true if $f:A\longrightarrow B$ is bijective).

In the second case: the hypothesis is "there is a partition of $A$ such that $\left\{f(A_j)\right\}_{j\in J}$ is a partition for $B$, with $f|_{A_j}$ bijection itself" isn't totally clear. Namely, what is the range of $f|_{A_j}$? Obviously, $f|_{A_j}: A_j\longrightarrow f(A_j)$ is OK.

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    I'm not sure I understand the problem of the first case, could you elaborate a bit more? – 2017-01-05
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    @user8469759, a partition is a famyly of *disjoint* sets. – 2017-01-05