Say whether the following function is injective, surjective, bijective, or none of the above (note: you can only select one option): Domain: A = {1,2,3} Codomain: B = {w, x, y, z) f = {(3,w), (2,z), (1,z)} O Injective O Surjective None O Bijective

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Chapter2: Second-order Linear Odes
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Need help with this foundations of mathematics homework problem. I had two math tutors from bartleby answer this problem and one said the answer was none and the other said the answer was injective. I don't know which answers are correct.

 

**Question:**

Determine whether the following function is injective, surjective, bijective, or none of the above (note: you can only select one option):

- **Domain:** A = {1, 2, 3}
- **Codomain:** B = {w, x, y, z}
- **Function:** f = {(3, w), (2, z), (1, z)}

**Options:**

- ○ Injective
- ○ Surjective
- ○ None
- ○ Bijective

**Explanation:**

To analyze the function \( f \), note the mappings:

- 3 maps to w
- 2 maps to z
- 1 maps to z

**Injective (One-to-One):**

A function is injective if every element of the domain maps to a unique element in the codomain. In this case, since both 2 and 1 map to z, the function is not injective.

**Surjective (Onto):**

A function is surjective if every element of the codomain is mapped by some element of the domain. Here, x and y in the codomain are not mapped by any element from the domain, so the function is not surjective.

**Bijective (One-to-One and Onto):**

A function is bijective if it is both injective and surjective. Since this function is neither, it is not bijective.

**Conclusion:**

The appropriate selection is "None" because the function is neither injective, surjective, nor bijective.
Transcribed Image Text:**Question:** Determine whether the following function is injective, surjective, bijective, or none of the above (note: you can only select one option): - **Domain:** A = {1, 2, 3} - **Codomain:** B = {w, x, y, z} - **Function:** f = {(3, w), (2, z), (1, z)} **Options:** - ○ Injective - ○ Surjective - ○ None - ○ Bijective **Explanation:** To analyze the function \( f \), note the mappings: - 3 maps to w - 2 maps to z - 1 maps to z **Injective (One-to-One):** A function is injective if every element of the domain maps to a unique element in the codomain. In this case, since both 2 and 1 map to z, the function is not injective. **Surjective (Onto):** A function is surjective if every element of the codomain is mapped by some element of the domain. Here, x and y in the codomain are not mapped by any element from the domain, so the function is not surjective. **Bijective (One-to-One and Onto):** A function is bijective if it is both injective and surjective. Since this function is neither, it is not bijective. **Conclusion:** The appropriate selection is "None" because the function is neither injective, surjective, nor bijective.
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