2nd April 2005

Munkres §22

Ex. 22.2.

(a). The map p: X → Y is continuous. Let U be a subspace of Y such that p^{−1}(U) ⊂ X is open. Then

f^{−1}(p^{−1}(U)) = (pf)^{−1}(U) = id_{Y}^{−1}(U) = U

is open because f is continuous. Thus p: X → Y is a quotient map.

(b). The map r: X → A is a quotient map by (a) because it has the inclusion map A ↪ X as right inverse.

Ex. 22.3. Let g: R → A be the continuous map f(x) = x × 0. Then q ∘ g is a quotient map, even a homeomorphism. If the compostion of two maps is quotient, then the last map is quotient; see [1]. Thus q is quotient.

The map q is not closed for \{x × \frac{1}{x} | x > 0\} is closed but q(A ∩ \{x × \frac{1}{x} | x > 0\} = (0, ∞) is not closed.

The map q is not open for R × (−1, ∞) is open but q(A ∩ (R × (−1, ∞))) = [0, ∞) is not open.

Ex. 22.5. Let U ⊂ A be open in A. Since A is open, U is open in X. Since p is open, p(U) = q(U) ⊂ p(A) is open in Y and also in p(A) because p(A) is open [Lma 16.2].

SupplEx. 22.3. Let G be a topological group and H a subgroup. Let φ_{G}: G × G → G be the map φ_{G}(x, y) = xy^{−1} and φ_{H}: H × H → H the corresponding map for the subgroup H. Since these maps are related by the commutative diagram

\xymatrix{G × G \ar[r]^{φ_{G}} & G \\ H × H \ar[r]^{φ_{H}} \ar@{^{(}->}[u] & H \ar@{^{(}->}[u]}

and φ_{G} is continuous, also φ_{H} is continuous [Thm 18.2]. Moreover, any subspace of a T_{1}-space is a T_{1}-space, so H is a T_{1}-space. Thus H is a topological group by Ex 22.1.

Now, by using [Thm 18.1.(2), Thm 19.5], we get

φ_{G}(\overline{H} × \overline{H}) = φ_{G}(\overline{H × H}) ⊂ \overline{φ_{G}(H × H)} = \overline{H}

which shows that \overline{H} is a subgroup of G. But then \overline{H} is a topological group as we have just shown.

SupplEx. 22.5. Let G be a topological group and H a subgroup of G.

(a). Left multiplication with α induces a commutative diagram

\xymatrix{G \ar[d]_{p} \ar[r]^{f_{α}} & G \ar[d]^{p} \\ G/H \ar[r]_{f_{α}/H} & G/H}

which shows [Thm 22.2] that f_{α}/H is ¸.

(b). The saturation gH = f_{g}(H) of the point g ∈ G is closed because H is closed and left multiplication f_{g} is a homeomorphism [SupplEx 22.4].

(c). The saturation UH = ⋃_{h∈H} Uh = ⋃ g_{h}(U) is open because U is open and right multiplication g_{h} is a homeomorphism [SupplEx 22.4].

(d). The space G/H is T_{1} by point (b). The map G × G →^{φ} G: (x, y) ↦ xy^{−1} is continuous [SupplEx 22.1] and it induces a commutative diagram

\xymatrix{G × G \ar[d]_{p×p} \ar[r]^{φ} & G \ar[d]^{p} \\ G/H × G/H \ar[r]_{φ/H} & G/H}

where p × p is a quotient map since it is open as we have just shown. (It is not true that the product of two quotient maps is a quotient map [Example 7, p. 143] but is is true that a product of two open maps is an open map.) This shows [Thm 22.2] that φ/H is continuous and hence [SupplEx 22.1] G/H is a topological group.

SupplEx. 22.7. Let G be a topological group.

(a). Any neighborhood of U of e contains a symmetric neighborhood V ⊂ U such that V V ⊂ U. By continuity of (x, y) → xy, there is a neighborhood W_{1} of e such that W_{1}W_{1} ⊂ U. By continuity of (x, y) → xy^{−1}, there is a neighborhood W_{2} of e such that W_{2}W_{2}^{−1} ⊂ W_{1}. (Any neighborhood of e contains a symmetric neighborhood of e.) Now V = W_{2}W_{2}^{−1} is a symmetric neighborhood of e and V V ⊂ W_{1}W_{1} ⊂ U.

(b). G is Hausdorff. Let x ≠ y. Since the set {xy^{−1}} is closed in G, there is a neighborhood U of e not containing xy^{−1}. Find a symmetric neighborhood V of e such that V V ⊂ U. Then V x ∩ V y = ∅. (If not, then gx = hy for some g, h ∈ V and xy^{−1} = g^{−1}h ∈ V^{−1}V = V V ⊂ U contradicts xy^{−1} ∉ U.)

(c). G is regular. Let A ⊂ G be a closed subspace and x a point outside A. Since e is outside the closed set xA^{−1}, a whole neighborhood U of e is disjoint from xA^{−1}. Find a symmetric neighborhood V of e such that V V ⊂ U. Then V x ∩ V A = ∅. (If not, then gx = ha for some g, h ∈ V and xA^{−1} ∋ xa^{−1} = g^{−1}h ∈ V^{−1}V = V V ⊂ U contradicts xA^{−1} ∩ U = ∅)

(d). G/H is regular for any closed subgroup H ⊂ G. The point xH is closed in G/H because the subspace xH is closed in G. Let x ∈ G be a point ouside a saturated closed subset A ⊂ G. Find a neighborhood V of e such that V x and V A are disjoint. Then also V xH ∩ V A = ∅ since A = AH is saturated. Since p: G → G/H is open [SupplEx 22.5.(c)], p(V x) is an open neighborhood of p(x) disjoint from the open neighborhood p(V A) of p(A). This shows that G/H is regular.

References

[1] Jesper M. Møller, General topology, http://www.math.ku.dk/~moller/e03/3gt/notes/gtnotes.pdf.