Math 205C - Topology Midterm

Math 205C - Topology Midterm Erin Pearse 1. a) State the definition of an n-dimensional topological (differentiable) manifold. An n-dimensional topological manifold is a topological space that is Hausdorff, has a countable basis at every point, and is locally Euclidean. That is, every point has a neighbourhood which is homeomorphic to an open set of R n. An n-dimensional differentiable manifold M is an n-dimensional topological manifold with a differentiable structure. That is, there is a collection of coordinate charts = {(U i, ϕ i )} which cover M such that i) x M, U x s.t. x U x and U x = V R n ii) For any two charts (U, ϕ) and (V, ψ), U V = ψ ϕ 1 and ϕ ψ 1 are diffeomorphisms of ϕ (U V ) and ψ (U V ) in R n. iii) U is maximal in the sense that any chart (U, ϕ) which compatible with in the sense of (ii) is included in. b) Describe all 1-dimensional manifolds. If a 1-dimensional manifold is compact, it is homeomorphic to S 1. If a 1- dimensional manifold is not compact, it is homeomorphic to R 1. Anything not falling into either category can readily be shown to be (i) not 1-dimensional, or (ii) not a topological manifold. c) Describe all closed orientable and non-orientable 2-dimensional manifolds. If a 2-dimensional closed manifold is orientable, then it is a sphere, a torus, or a connected sum of tori. That is, it is an n-genus torus (with a sphere corresponding to genus 0). If a 2-dimensional closed manifold is non-orientable, then it is a Klein bottle, projective plane, or connected sum of them. d) What are the fundamental groups of the manifolds in (b) and (c)? π 1 (S 1 ) = Z and S 1 has universal covering space R 1. π 1 (R 1 ) = {1} and R 1 is its own universal covering space. π 1 (S 2 ) = {1} and S 2 has universal covering space R 2. π 1 (T 2 ) = Z Z and T 2 has universal covering space R 2. π 1 (# g i=1 T 2 ) = Z 2g and # g i=1 T 2 has universal covering space R 2. π 1 (K) = F (a,b) and K has universal covering space {aba 1 b} R2. ( π ) 1 RP 2 = Z2 and RP 2 has universal covering space S 2.

2 Math 205C - Topology Midterm Erin Pearse 2. Explain why each of the following figures either is or is not a topological manifold. a) This object is clearly 1-dimensional. Take x to be an intersection point with open nbd U as shown. Assuming this space has the natural (subspace) topology, U would have to be homeomorphic to an open set V R 1. Since U {x} consists of three connected components, and for V {f(x)} consists of only two, f : U V cannot be a homeomorphism. This is explained in a little better detail in (v), below. b) By the same argument as above, any open nbd of the intersection point x cannot be homemorphic to an open set of R 1. c) Though not a differentiable manifold, this space is clearly homeomorphic to (a, b) R 1 and is thus a 1-dimensional topological manifold. d) This space is not Hausdorff, as any open set containing x also contains y. Since it is not Hausdorff, it fails to be a topological manifold. e) Any point of this space has a nbd homeomorphic to an open subset V R 2, unless it lies in the intersection of the two planes. In this case, an open nbd U of x looks like two intersecting plates. Let l be the line that forms the intersection, and suppose f : U V R 2 is a homeomorphism. But U\l consists of 4 distinct connected components, and f(u\l) can consist of at most two connected components. To see this, note that for f(l) to separate R 2 into more components, f l would not be injective. Thus, this space is not a topological manifold.

Math 205C - Topology Midterm Erin Pearse 3 3. a) Prove that R 1 cannot be homeomorphic to R 2, for n 2. Let f : R 1 R n, and let U = R 1 {x} where x R 1. Then if f were a homeomorphism, f R {x} : R {x} R n would also be a homeomorphism. Then, R {x} is disconnected f (R {x}) is disconnected. But for n 2, R n will always be connected! < Therefore, such a homeomorphism cannot exist. b) Try to prove that R n cannot be homeomorphic to R m, for n m. Begin by taking n < m. Then we imbed f : R n R m by inclusion. Note that any open set of R m must have dimension m. f(r n ) = R n has dimension n, so clearly R n cannot be an open subset of R m. Thus, f cannot be a homeomorphism, because it takes an open set to a non-open set. 1 4. a) Give an example of a closed 3-dimensional manifold whose fundamental group is (Z, +). i) S 2 S 1 b) Give two examples of a closed simply-connected 4-dimensional manifold. i) S 2 S 2 ii) S 4 c) Give two examples of a closed, simply-connected 3-dimensional manifold. i) S 3 ii) Umm... 1 I understand why this argument is wrong - the result is due to a confusion of the topologies on R n relative to the homeomorphism and to R m. However, this problem asks us to try to prove the theorem, and this was the best I could come up with, short of re-deriving Brouwer s Invariance of Domain from scratch.

4 Math 205C - Topology Midterm Erin Pearse 5. Let M be a topological (differentiable) manifold with an infinite fundamental group. Prove that its universal covering M is a non-compact topological (differentiable) manifold. Since M is a universal covering, it is path-connected, so for any point m 0 M, we can use the lifting correspondence to establish a bijection π 1 (M) Φ : ( ) p 1 (m 0 ) π 1 M (cf. [Munk] Thm. 54.6b). Then we immediately have that p 1 (m 0 ) π 1 (M) = ( ) by bijection π 1 M = π 1 (M) {1} = π 1 (M) M is simply connected = by hypothesis Now let U be an open nbd of m 0 so that m 0 U M. Because p is a covering map, we have p 1 (U) = i I V i where I is an infinite indexing set and p Vi : V i U is a homeomorphism. Let {x i } i=1 by any countable sequence of points in p 1 (m 0 ). Then {x i } is an infinite sequence in M, and x i V i i. Since the definition of covering map guarantees that all the V i are disjoint, {x i } clearly cannot have any limit points. Thus, by the equivalence of compactness and limit point compactness in metrizable spaces, M must not be compact.

Math 205C - Topology Midterm Erin Pearse 5 6. Define the following terms: a) rank of a differentiable mapping The rank of a differentiable mapping F at p is the rank at a = ϕ(p) of the Jacobian matrix 1 x 1 1 x n f a =..... m x 1 b) immersion An immersion is a differentiable mapping f : M N such that dim M = rank f at every point of M. In this case, f : T p (M) T ϕ(p) (M) is injective. c) submersion An submersion is a differentiable mapping f : M N such that dim N = rank f at every point of N. In this case, f : T p (M) T ϕ(p) (M) is surjective. d) imbedding An imbedding is an injective immersion f : M N which is homeomorphic to its image f(m) N. In this case, the topology on f(m) induced by f and the subspace topology on f(m) induced by N are identical. e) submanifold (i.e., immersed submanifold) A submanifold or immersed submanifold is just the image of an immersion, that is, the image of a manifold M under an immersion. In this case, the topology on f(m) is taken as being induced by the topology on M, via the immersive map. f) imbedded submanifold An imbedded submanifold is the image of an imbedding, that is, the image of a manifold under an immersion. m x n a

6 Math 205C - Topology Midterm Erin Pearse 7. a) Is an injective immersion an imbedding? If not, give two examples. i) Define the figure-eight map G : R R 2 as G (t) = ( 2 cos ( π + 2 arctan t), sin 2 ( π + 2 arctan t)) 2 2 G 1 (t) will not be continuous at (0, 0) for reasons similar to those discussed in #2(b,e). The essential ideal is that any sufficiently small neighbourhood of (0, 0) which is open in the subspace topology of R 2, will be mapped by G 1 (t) to a disjoint union of three subsets of R. ii) Define a variant of the topologist s sine curve as ( 1, sin πt) t > 1 t 2 F (t) = (smooth) 0 t 1 2 (0, t + 3) t < 0 where smooth is a smooth arc connecting the other two sections of the graph such that the entire set is C. (see [Boot] III.4.10, Fig. III.8) The idea of this example is similar to the previous one. F 1 (t) will not be continuous at any point in the set {(0, y) R 2. 1 < y < 1}. b) Let f : M N be a injective immersion. Is f(m) a closed subset in N? If not, give an example. No, f(m) need not be closed in N. Define f : R R 2 by f (x) = ( 2 arctan x, 0) π so that f (R) = {(x, 0). 1 < x < 1} Since the limit points ( 1, 0) and (1, 0) are not contained in f (R), f (R) must not be closed. c) Let f : M N be an imbedding. Is f(m) a closed subset in N? If not, give an example. No, f(m) need not be closed in N. Define f : R 2 R 2 by f (x, y) = ( 2 arctan x, 2 arctan y) π π so that f ( R 2) = {(x, y) R 2. 1 < x, y < 1} Then f (R 2 ) is an open rectangle in R 2 and clearly not closed.

Math 205C - Topology Midterm Erin Pearse 7 d) Let f : M N be an injective immersion, and consider f(m) as endowed with the subspace topology from N. i) Is f : M f (M) a continuous map? An immersion is by definition differentiable, and continuity is a prerequisite for differentiability. So f is clearly continuous onto its image. ii) Is f 1 : f (M) M a continuous map? No, not in general. The injective immersion f : R F of R as a figureeight F R 2 is a counterexample. Let the intersection point of F be at the origin, and let B ε (0) be the ball of radius ε centered at the origin. For small enough ε, it is clear that any point x on two of the four branches of B ε (0) F, f 1 will take x outside of ( 1, 1). In other words, f 1 is not continuous at 0 R 2. iii) What happens when M is a compact manifold? When M is compact, we can show that f must be proper, from which it follows (by #8.b) that f is an imbedding, and that f(m) is a regular submanifold. Let K M be a closed subset of M. Then we have K closed K compact (M is compact) f(k) compact (f is continuous) f(k) closed (N is Hausdorff) This shows that f is a closed map. Now let C be a compact subset of N, and we will show f 1 (C) is compact. Let {U α } α A be an open covering of f 1 (C). For any y C, α such that f 1 (y) U α. Denote by U y the U α that contains f 1 (y). Now define V y = N f (M U y ) and note that V y is an open set containing y. V y is an open because f (and hence f (M U y )) is closed. Now C is compact, and {V y } covers C, so we can find a finite subcover {V yi } n i=1. Then f 1 (C) f 1 (V y1 )... f 1 (V yn ) because U y1... U yn f 1 (V y ) = f 1 (Y f (X U y )) f 1 (Y ) (X U y ) = X (X U y ) = U y f 1 (V y ) U y. But then {U yi } n i=1 is a finite subcover of f 1 (C), so f 1 (C) is compact. Hence, f is proper. (see also [Bred] I.7.13 and [Boot] III.5.7)

8 Math 205C - Topology Midterm Erin Pearse 8. a) Let f : M N be a proper continuous map between two topological manifolds M and N. Prove that f(m) is a closed subset of N. Since M,N are manifolds, they are locally compact. Let Ṁ denote the 1-point compactification of M. Note that if M is already compact, then the new point m is an isolated point of M that is clopen, i.e., both closed and open. 2 Similarly, take (Ṅ = N n) to be the 1-point compactification of M. Define f ( m ) = n and note that this extends f : Ṁ Ṅ continuously as follows: For U open in N, f 1 (U) is open in M by continuity. For U = Ṁ K, we have f 1 (U) = f 1 1 (Ṅ K) = f (Ṅ) f 1 (K) = Ṁ f 1 (K), which is open in Ṁ by proper. Now we have that Ṁ is compact by construction, so f(ṁ) is compact as the continuous image of a compact set. Since N, and hence Ṅ, is Hausdorff, this gives us that f(ṁ) is closed in Ṅ as a compact subset of a Hausdorff space. But N is closed in Ṅ, so then f(ṁ) N is closed in N by subspace topology. But f(ṁ) = f (M m) = f (M) f ( m ) = f (M) n. Thus, f(ṁ) N = (f (M) n) N = f (M). Hence f(m) is closed in N. Note: for D closed in M, D Ṁ implies that f(d) is compact, and hence closed in Ṅ. So this proof actually shows that f is a closed map! b) Let f : M N be a proper injective immersion of differentiable manifolds M and N. Prove that f is an imbedding and f(m) is a regular submanifold and a closed subset of M, and conversely. At the end of part (a), it was shown that f is a closed map. f is given to be injective, so f is a bijection when restricted to its image as f : M f (M). Then f : M f (M) is also an open map, which is equivalent to saying f 1 is continuous. Then f a bijection and f, f 1 continuous shows that f : M f (M) is a homeomorphism and hence that f : M N is an imbedding. Note that f(m) is closed in N by part (a), and that ([Boot] III.5.5, p.78) implies that f(m) is regular. f is given to be an imedding, so we immediately have that f is an injective immersion, by definition. To see that f is proper, choose K N compact. Then N Hausdorff implies K is closed. Since f(m) is closed by (a), this shows K f (M) is a closed subset of K, and hence is also compact. Now f is an imbedding, so f 1 is continuous. Thus, f 1 (K) is a continuous image of a compact set, and is hence compact. So we have shown K compact f 1 (K) compact. 2 This follows from the definition of the topology of Ṁ: U is open in Ṁ iff U is open in M or U = Ṁ K for some compact set K M.

Math 205C - Topology Midterm Erin Pearse 9 9. Let f : R 3 R 4 be a mapping given by f (x 1, x 2, x 3 ) = (x 1 x 2, x 1 x 3, x 2 x 3, x 1 + x 2 + x 3 ). a) Compute the Jacobian of f at (1, 1, 1). 1 1 1 x 1 x 2 x 3 x 2 x 1 0 x = 2 2 2 x 1 x 2 x 3 x 3 0 x 1 3 3 3 = x 1 x 2 x 3 0 x 3 x 2 1 1 1 x = (1,1,1) = 4 4 4 x 1 x 2 x 3 1 1 0 1 0 1 0 1 1 1 1 1 0 0 0 0 0 1 0 1 0 1 0 0 = = 0 1 1 1 0 1 0 1 1 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 = 0 0 2 0 0 1 0 1 1 1 0 0 b) What is the rank of f at (1, 1, 1)? ( By the above calculation, rank x ) (1,1,1) = 3. c) Could you find a point of R 3 such that the rank of f is less than 3? On the x 1 -axis, when x 2 and x 3 are 0, we have: 0 a 0 x = 0 0 a (a,0,0) 0 0 0, 1 1 1 which still has rank 3 for any a. Hence, we will only have rank less than three at the origin: 0 0 0 x = 0 0 0 (0,0,0) 0 0 0, 1 1 1 which has rank 1. d) What are the points in R 3 at which f is immersive? f is immersive on R 3 \{0}.

10 Math 205C - Topology Midterm Erin Pearse 10. State and understand the Inverse Function Theorem, the Implicit Function Theorem, and the Rank Theorem. a) The Inverse Function Theorem. Let W be an open subset of R n and f : W R n a C r mapping for r Z +. If a W and f (a) is nonsingular, then there exists an open neighbourhood U of a in W such that V = f(u) is open and f : U V is a C r diffeomorphism. If x U and y = f(x), then the derivatives of f 1 at y are given by ( f 1 ) (y) = (f (x)) 1 where (f (x)) 1 is the inverse matrix of f (x). b) The Implicit Function Theorem. (Surjective form). Let U R m be an open set and f : U R n a C r map, r Z +. Let p U, f(p) = 0, and suppose that f x is surjective. Then there exists a local diffeomorphism ϕ of R m at 0 such that ϕ(0) = p and f ϕ (x 1,..., x m ) = (x 1,..., x n ). (Injective form). Let U R m be an open set and f : U R n a C r map, r Z +. Let q R n be such that 0 f 1 (q), and suppose that f 0 injective. Then there exists a local diffeomorphism ψ of R n at q such that ψ(q) = 0 and ψ f(x) = (x 1,..., x m, 0,..., 0). c) The Rank Theorem. Let A 0 R n, B 0 R m be open sets, f : A 0 B 0 be a C r mapping, and suppose the rank of f on A 0 to be equal to k. If a A 0 and b = f(a), then there exist open sets A A 0 and B B 0 with a A and b B. Also, there exist open sets U R n and V R m and C r diffeomorphisms g : A U, h : B V such that h f g 1 (U) V and such that h f g 1 (x 1, x 2,..., x n ) = (x 1, x 2,..., x n, 0,..., 0).

Math 205C - Topology Midterm Erin Pearse 11 11. Let M and N be two n-dimensional differentiable manifolds. a) Suppose f : M N is an immersion. Prove that f is an open map. Let W M be open and pick y f(w ). Then y = f(x) for some x W. Since f is an immersion, we know Df(x) is nonsingular. By the Inverse Function Theorem, x has an open neighbourhood U W such that f (U) is open. Then U W implies f(u) f(w ) is an open neighbourhood of y. Since we can do this for any y, f(w ) is open. Alternative proof: Let W M be open and consider f(w ). Pick y f(w ) so y = f(x) for some x W. Then x has a neighbourhood U such that f U is an imbedding of U in N, by ([Boot] III.4.12, p.74). Thus, f(u) f(w ) is an open neighbourhood of y. Since we can do this for any y, f(w ) is open. b) Suppose f : M N is an immersion and M is compact. Prove that f is a surjection. M is open in M trivially, so f(m) is open in N by (a). Then M is compact by hypothesis, so we know f(m) is compact and hence closed, because N is Hausdorff. Evidently f(m) is both open and closed. Observe that N is connected. 3 Since the only clopen subsets of a connected space are the entire space and the empty set, it must be the case that f(m) = N. 3 At this point, we use the fact that we are considering all manifolds to be connected. If this condition is not included in the definition, this theorem can readily be demonstrated to be false. For example, let M be S 1 and let N be a disjoint union of two copies of S 1. Then any immersion of M in N will not be surjective.

12 Math 205C - Topology Midterm Erin Pearse 12. a) What is a C partition of unity? A C partition of unity on M is a collection of C functions {f α } defined on M and satisfying: i) f α 0 on M ii) {supp (f α )} forms a locally finite covering of M iii) α f α (x) = 1, x M b) Let p q be two distinct points on a differentiable manifold M. Prove that there exists a C function f : M R such that f(p) = 100 and f(q) = 1001. Let g be the imbedding of M into R n, as guaranteed by the Whitney Imbedding Theorem. Then let b c, ε (x) : R n R be the bump function centered at c with support {x R n. x c ε}, as defined in class. Then set ε = 1 g(p) g(q) 2 and define b as follows: b (x) = 100b p,ε (x) + 1001b q,ε (x) Finally, we define f = b g. Because C (R n ) is an algebra, f is clearly C. Alternative proof: As a manifold, M is Hausdorff, so we can find two disjoint open sets U and V such that p U and q V. Then since M is locally compact, we can find a neighbourhood C of x such that C is compact and C U, and we can find a neighbourhood D of y such that D is compact and D V. By ([Boot] III.3.4, P.67) there exist C functions 0, x M\U 0, x M\V b C (x) = 0 < α < 1, x U\C and b D (x) = 0 < β < 1, x V \D. 1, x C 1, x D Now define f (x) = 100b C (x) + 1001b D (x).

Math 205C - Topology Midterm Erin Pearse 13 13. a) State the Whitney Imbedding Theorem The easy Whitney Imbedding Theorem states: Any differentiable manifold M of dimension n may be imbedded differentiably as a closed submanifold of R 2n+1. The hard Whitney Imbedding Theorem states: For n > 0, every paracompact Hausdorff n-manifold can be imbedded into R 2n. Furthermore, it may be immersed in R 2n 1 if n > 1. b) Let M be a compact differentiable manifold of dimension n. Prove that there exists an imbedding f : M R n for sufficiently large n, without using (a). Take n = dim M and let {A α } be any open covering of M. By ([Boot] V.4.1), there exists a countable, locally finite refinement {U i } i=1 consisting of coordinate neighbourhoods (U i, ϕ i ) with i) ϕ i (U i ) = B n 3 (0) i, and ii) for V i = ϕ 1 i (B1 n (0)) U i, M = i=1 V i Since M is compact, we need consider only a finite set of coordinate charts {(U i, V i, ϕ i )} k i=1. Now by ([Boot] V.4.4), we can take a subordinate C partition of unity {g i } k i=1 such that g i = 1 on V i. Define the C maps f i : M R n by { g i ϕ i, x U i f i (x) = 0, otherwise so that each f i is immersive on V i. Now for R N = R n+1 R n+1 = R (n+1)k and F i = (f i, g i ), define F : M R N by F (p) = (F 1 (p),..., F k (p)) F is componentwise C and thus C. F is injective. Pick x y, with y V i. case i) x V i. Then f i Vi = ϕ i Vi = F (x) F (y) (ϕ i is injective) case ii) x / V i. Then g i (y) = 1 g i (x) = F (x) F (y) F is an immersion. For x M, x V i for some i. Then = g i, and hence g, is immersive at x, as in the proof of #16. Now we can apply #7(d)iii to obtain the desired result. (See [Boot], p.196)

14 Math 205C - Topology Midterm Erin Pearse 14. Let M be a differentiable manifold. Prove that there exists a proper differentiable map f : M R n, for any positive integer n. Pick some point x 0 M and define f(x) = d(x 0, x), where d is the Riemannian metric. This is justified, because it is possible to define a C Riemannian metric on every C manifold (cf. [Boot] V.4.5, p.195), and because a Riemannian manifold (that is, manifold on which a Riemannian metric Φ is defined) is a complete metric space in which the metric topology and manifold topology coincide (cf. [Boot] V.3.1, p.189). f is given explicitly by f (x) = inf c(t) D 1 { v u ( Φ ( dp dt, dp dt )) 1/2 dt. c (u) = x 0, c (v) = x, u t v} Then f : M R + is a metric and hence differentiable. For any compact interval [a, b] R +, its preimage will be an annulus about x 0 (i.e., f 1 ([a, b]) = {x R. a x x 0 b} ). This is enough to show that the preimage of any compact set is compact. Since f : M R is proper and differentiable, we can define the imbedding g : R R n by inclusion. Then g f will work, for any n. Alternative proof: To avoid the use of Riemannian metrics, this proof can be adapted as follows: use the Whitney embedding theorem to embed M in some R N, then replace Φ with the standard n-dimensional Euclidean metric. The rest of the proof remains essentially the same.

Math 205C - Topology Midterm Erin Pearse 15 15. Prove that there exists no immersion f : S n R n. S n is compact and f is continuous, so f(s n ) is compact. Since R n is Hausdorff, f(s n ) is also closed in R n. Following ([Munk] p.102), we define the boundary of a set A X by A = A (X A) Then we can pick a point x (S n ) and then also a point y f 1 (x) f 1 ( (S n )). Now, if f really were an immersion, the Df would be nonsingular everywhere. In particular, Df would be nonsingular at x. Then we can find an open nbd U S n of x such that V = f(u) is open in R n by the Inverse Function Theorem. Yet f(u) is clearly not open, because x f(u) and any open nbd of x is not contained in f(u) by the choice x (S n ). To see this, note that the definition of boundary makes x a limit point of (R n f (S n )). Alternative proof: Note that S n and R n are manifolds of the same dimension, and S n is compact. Then we know by #11 that any immersion f : S n R n would be a surjection and that f is an open map. Let {U α } a A be an open covering of R n. Then {f 1 (U α )} a A is an open covering of S n, so it must have a finite subcover {f 1 (U αi )} n i=1. Thus, {f (f 1 (U αi ))} n i=1 = {U α i } n i=1 (equality because f is surjective) shows that any open covering of R n has a finite subcovering. < (R n is not compact)

16 Math 205C - Topology Midterm Erin Pearse 16. Let f : M R n be an injective immersion. Prove that there exists an imbedding F : M R n+1 such that f(m) is a closed subset of R n+1. We have f (x) = (f 1 (x), f 2 (x),..., f n (x)), and we know from the proof of #14 that we can find a proper differentiable map g : M R. Define F (x) = (f 1 (x), f 2 (x),..., f n (x), g (x)) So that F : M R n+1. We will show that F is a proper injective immersion. Then the result will follow by #8(b). F is proper. Let {x i } i=1 be any divergent sequence in M. Then {g (x i)} i=1 will be a divergent sequence because g is proper. Then it must also be the case that {F (x i )} i=1 = {(f 1 (x i ),..., f n (x i ), g (x i ))} i=1 is a divergent sequence. This shows that F is proper. F is injective. Let m 1 m 2 be distinct points of M. Then f is injective, so f (m 1 ) f (m 2 ). This implies F (m 1 ) = (f (m 1 ), g (m 1 )) (f (m 2 ), g (m 2 )) = F (m 2 ). F is an immersion. First we note that f : M N is an immersion Df : T p (M) T f(p) (N) is injective. ( ) Pick v T p (M) such that DF (v) = 0. Then 1 x 1 1 x m DF (v) =... v 1.. n x 1 n. = x m g g v m x 1 x m This can only be true if 1 v dx Df (v) = n dx. v = 1 v dx. n v dx g v dx = And since f is an immersion, we know by ( ) that Df is injective. Hence, this can only happen when v = [0,..., 0]. This shows that 0. 0 DF (v) = 0 v = 0, i.e., that DF is injective. Then by ( ) again, F is an immersion. 0. 0 0

Math 205C - Topology Midterm Erin Pearse 17 Proof of ( ) f : M N is an immersion Df : T p (M) T f(p) (N) is injective. Note that dim M = dim T p (M) and dim V = rank A + dim (ker A), for any linear mapping A defined on a vector space V. Since Df : T p (M) T f(p) (N) is linear, this gives Now we have dim T p (M) = rank Df + dim (ker Df) f : M N is an immersion dim M = rank Df dim (ker Df) = 0 ker Df = {0} Df is injective.

18 Math 205C - Topology Midterm Erin Pearse References [Bred] [Boot] [Hirs] [Mass] [Miln] [Munk] Bredon, Glen E. (1993) Topology and Geometry. Springer-Verlag. Boothby, William M. (1986) An Introduction to Differentiable Manifolds and Riemannian Geometry (Second Edition). Academic Press. Hirsch, Morris W. (1976) Differential Topology. Springer-Verlag. Massey, William S. (1967) Algebraic Topology: An Introduction. Springer-Verlag. Milnor, John W. (1965) Topology from the Differentiable Viewpoint. Princeton University Press. Munkres, James R. (2000) Topology (Second Edition). Prentice Hall.