Yes, projection maps are always surjective.
A projection map is a fundamental operation in set theory and various branches of mathematics, including topology and linear algebra. It operates on elements of a Cartesian product.
As defined in the provided reference, a projection map, often written as projj, takes an element from a Cartesian product of sets, say (X_1 \times \cdots \times X_j \times \cdots \times X_n), and returns its component in the j-th position.
Specifically, if you have an element (x = (x_1, \dots, x_j, \dots, x_n)) in the Cartesian product space, the projection map onto the j-th component, projj, is given by:
( \text{proj}_j(x) = x_j )
Here, (x_j) is the element from the set (X_j).
Are Projection Maps Surjective?
Based directly on the definition and the provided reference, the answer is a definitive yes.
The reference explicitly states: "This map is always surjective".
Why are Projection Maps Surjective?
A function is surjective (or "onto") if for every element in the codomain, there is at least one element in the domain that maps to it.
Let's consider the projection map ( \text{proj}_j: X_1 \times \cdots \times X_n \to X_j ).
The codomain of this map is the set (X_j). For the map to be surjective, for any element (y) that you pick from the set (X_j), there must exist some element (x = (x_1, \dots, x_n)) in the Cartesian product (X_1 \times \cdots \times X_n) such that ( \text{proj}_j(x) = y ).
We can easily construct such an (x). Let (y) be any element in (X_j). We can choose any elements (x_1 \in X1, \dots, x{j-1} \in X{j-1}, x{j+1} \in X_{j+1}, \dots, x_n \in X_n) (assuming these sets are non-empty – if any (X_k) is empty for (k \neq j), the product space is empty, making the question trivial or ill-posed depending on convention; assuming standard non-empty sets). Then, we form the tuple (x = (x1, \dots, x{j-1}, y, x_{j+1}, \dots, x_n)). This tuple (x) is an element of the Cartesian product (X_1 \times \cdots \times X_n).
When we apply the projection map ( \text{proj}_j ) to this (x), we get:
( \text{proj}_j(x) = \text{proj}_j(x1, \dots, x{j-1}, y, x_{j+1}, \dots, x_n) = y )
Since we were able to find an element (x) in the domain ((X_1 \times \cdots \times X_n)) that maps to an arbitrary element (y) in the codomain ((X_j)), the projection map ( \text{proj}_j ) is indeed surjective.
Example
Consider the projection map from ( \mathbb{R}^2 ) (the Cartesian plane) onto the first component ((x)-axis).
Let ( \text{proj}_1: \mathbb{R}^2 \to \mathbb{R} ) be defined by ( \text{proj}_1((a, b)) = a ).
Is this map surjective? The codomain is ( \mathbb{R} ). We need to check if for every real number (y \in \mathbb{R}), there is a point ((a, b) \in \mathbb{R}^2) such that ( \text{proj}_1((a, b)) = y ).
Yes, we can always find such a point. For any (y \in \mathbb{R}), we can choose the point ((y, 0)) in ( \mathbb{R}^2 ). Applying the projection:
( \text{proj}_1((y, 0)) = y )
Since we can find a point ((y, 0)) in ( \mathbb{R}^2 ) for any (y \in \mathbb{R}) that maps to (y), the map ( \text{proj}_1 ) is surjective. We could have chosen any other second coordinate, like ((y, 5)) or ((y, -100)).
Key Takeaway
The core property that makes projection maps surjective is that for any desired output element (y) from the target set (X_j), you can construct an input tuple in the Cartesian product simply by placing (y) in the j-th position and filling the other positions with any valid elements from their respective sets.
