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    Collapsing the disk fibers of $S^1\times D^2$ yields $S^1$

    P-T09A3

    Area
    Topology
    Topics
    Homeomorphisms, Quotient Spaces
    Seen at
    UNL
    Years
    2009
    Status
    draft

    Let \(S^1\) and \(D^2\) be the unit circle and closed unit disk in the (Euclidean) space \(\mathbb R^2\). Let \(\sim\) be the smallest equivalence relation on the product space \(S^1\times D^2\) satisfying \((x,y)\sim(x,(0,0))\) for all \(x\in S^1\) and \(y\in D^2\). Let \(Z:=(S^1\times D^2)/\sim\) be the corresponding quotient space. Prove that \(Z\) is homeomorphic to \(S^1\).

    Solution

    Let \[ \pi:S^1\times D^2\longrightarrow S^1, \qquad \pi(x,y)=x, \] and let \[ q:S^1\times D^2\longrightarrow Z \] be the quotient map.

    1. The equivalence classes of \(\sim\) are exactly the sets\[ \{x\}\times D^2 \qquad(x\in S^1). \]

    Each generating relation has the form \[ (x,y)\sim(x,(0,0)), \] so taking the equivalence relation generated by these pairs never changes the first coordinate. Thus equivalent points have the same first coordinate.

    Conversely, if \((x,y)\) and \((x,y')\) have the same first coordinate, then \[ (x,y)\sim(x,(0,0))\sim(x,y'), \] where the second relation follows by symmetry of an equivalence relation. Hence all points of \(\{x\}\times D^2\) are equivalent.

    2. There is a unique continuous map\[ f:Z\longrightarrow S^1 \]such that\[ f(q(x,y))=x. \]

    The projection \(\pi\) is continuous and, by 1, constant on every equivalence class of \(\sim\). By the universal property of the quotient topology, it therefore factors uniquely through \(q\) as \[ \pi=f\circ q \] with \(f\) continuous.

    3. The map \(f:Z\to S^1\) is bijective.

    For surjectivity, if \(x\in S^1\), then \[ f(q(x,(0,0)))=x. \]

    For injectivity, suppose \[ f(q(x_1,y_1))=f(q(x_2,y_2)). \] Then \(x_1=x_2\), so 1 gives \[ (x_1,y_1)\sim(x_2,y_2). \] Thus their quotient classes agree: \[ q(x_1,y_1)=q(x_2,y_2). \]

    4. The quotient space \(Z\) is compact.

    Both \(S^1\) and \(D^2\) are compact. The finite product theorem therefore gives compactness of \[ S^1\times D^2. \] The quotient map \(q\) is continuous and surjective, so its image \[ Z=q(S^1\times D^2) \] is compact.

    5. The map \(f\) is a homeomorphism, and hence\[ Z\cong S^1. \]

    By <1>2–<1>3, \(f\) is a continuous bijection. Its domain \(Z\) is compact by 4, and its target \(S^1\), being a subspace of the Hausdorff space \(\mathbb R^2\), is Hausdorff. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.