- #1

- 51

- 0

x | f(A)=B | |

x | In the arrow diagram representing f, every point in B has an arrow pointing at it. | |

x | ∀y∈B ∃x∈A such that f(x)=y | |

x | f−1(B)=A | |

x | Every element of B is the image of some element in A | |

∀x∈A ∃y∈B such that f(x)=y | ||

Every element of A has a corresponding image in B | ||

x | Every element of B has at least one preimage. |

I have marked with an "x" those that I believe are correct. However, I am not entirely sure.

Let me explain my selection.

1. f(A) = B is true, because B is the whole set B where A is mapped to B, so it is onto? (range = codomain)

2. Valid, because it shows that the range = codomain.

3. Similar to point 2, it shows that range = codomain.

4. In this case, it is similar to 1? It seems to say that the whole set of B can be mapped back to A.

5. Yes, every element of B (codomain) is the image of some x in A.

6. Not true for onto function, because it does not cover all of B, only some in B.

7. Not true for onto function, because it does not talk about covering all options in B.

8. True because every element of B is covered.

Please point out if I have made any mistakes in my explanation, thank you!