Prove that the projection map is closed when a subset C is closed in a quotient space | Step-by-Step Solution

Understanding Quotient Space Projections

What We're Solving:

We need to prove that the projection map π: X → X/C is closed, where C is a closed subset of X, and our equivalence relation says two points are equivalent if they're the same point OR if they're both in C.

The Approach:

This is a beautiful problem because it connects the geometric intuition of "collapsing" a closed set to a point with the rigorous definition of closed maps! Our strategy is to take any closed set in X and show its image under π is closed in X/C. We'll use the characterization that a set is closed if and only if its complement is open.

Step-by-Step Solution:

Step 1: Understand the quotient space structure

• In X/C, every point in C gets "collapsed" to a single equivalence class [C]
• Points outside C remain as individual equivalence classes [x] where x ∉ C
• So X/C looks like: {[C]} ∪ {[x] : x ∉ C}

Step 2: Set up what we need to prove To show π is closed, we need: for any closed set F ⊆ X, prove π(F) is closed in X/C.

Step 3: Use the complement approach Instead of directly proving π(F) is closed, let's prove its complement is open! We need to show: (X/C) \ π(F) is open in X/C.

Step 4: Analyze the complement (X/C) \ π(F) = {[x] ∈ X/C : [x] ∩ F = ∅}

This means we're looking at equivalence classes that don't intersect F.

Step 5: Connect back to X An equivalence class [x] doesn't intersect F if and only if:

• If x ∉ C, then x ∉ F
• If x ∈ C, then C ∩ F = ∅ (since [x] = [C] when x ∈ C)

Step 6: Express in terms of preimage (X/C) \ π(F) = π(X \ F) when we're careful about the equivalence classes!

The preimage π⁻¹((X/C) \ π(F)) = X \ F.

Step 7: Use the key property Since F is closed in X, we know X \ F is open in X. Since π is the quotient map, a set U in X/C is open if and only if π⁻¹(U) is open in X.

Step 8: Conclude Since π⁻¹((X/C) \ π(F)) = X \ F is open in X, we conclude that (X/C) \ π(F) is open in X/C.

Therefore, π(F) is closed in X/C.

The Answer:

The projection map π: X → X/C is closed. The proof relies on showing that for any closed set F in X, the set π(F) has an open complement in X/C, which we established by using the quotient topology definition and the fact that X \ F is open when F is closed.

Memory Tip:

Remember: "Quotient maps preserve the closed/open nature when you trace through complements!" The key is always to use the defining property of quotient topology: U is open in X/C ⟺ π⁻¹(U) is open in X.

Great job working through this topology problem! The interplay between equivalence relations, quotient spaces, and continuity properties is one of the most elegant parts of topology. 🌟