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i can't seem to make headway in finding a function that satisfies the conditions as shown in the picture](https://i.stack.imgur.com/nlR0o.jpg)

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    Interesting question! What have you tried so far? – 2017-02-24
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    I used the fact that f1•f is bijective which implies f is injective and the fact that f•f2 is bijective which implies f is surjective, which overall means f is bijective which would imply A~B. But after that I don't know where to look to find f – 2017-02-24
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    Good points. Do you think maybe those three are separate problems? I don't think you can find one function that satisfies all the conditions in (1), (2), *and* (3). – 2017-02-24
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    They very well could be, I don't know why I assumed that they were all part of the same question! – 2017-02-24

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Your first comment (which should have been part of your question!) is the key: No single function can satisfy all three of these conditions. In fact, no function can satisfy the first two.

If $f_1 \circ f = \operatorname{id}_A$, then $f$ is an injection and $f_1$ is a surjection. If $f \circ f_2 = \operatorname{id}_B$, then $f_2$ is an injection and $f$ is a surjection. So if both (1) and (2) are true, $f$ must be a bijection. Then we can compose both sides of $f\circ f_2 = \operatorname{id}_B$ by $f^{-1}$ to get $f_2 = f^{-1}$, which would tell us that $f_2 \circ f$ must be $\operatorname{id}_A$.

So that made me think that the problem was three separate ones grouped together. It's too bad that the preamble says "Find a function..."; perhaps "Find functions..." would have been more understandable.

To answer them separately:

  1. We know $f$ must be an injection and not a surjection. Perhaps the simplest case is to let $A = \{0\}$, $B = \{1,2\}$, $f(0) = 1$, $f_1(1) = f_1(2) = 0$. Then $(f_1 \circ f)(0) = 0$ but $(f \circ f_1)(2) = 1$.

  2. We know $f$ must be a surjection and not an injection. So let $A = \{0,1\}$, $B = \{2\}$, $f(0) = f(1) = 2$, and $f_2(2) = 0$. Then $(f\circ f_2)(2) = 2$, but $(f_2 \circ f)(1) = 0$.

  3. Let $f_3 = f_2$, and $f_4(2) = 1$.