User:NakkLeVaar

Differential Geometry

 * Implicit Function Theorem (I). Consider a $$C^1$$-differentiable map $$f:\R^n\times\R^m\rightarrow\R^m$$$$(x_1,\cdots,x_n,y_1,\cdots,y_m)\longmapsto(f_1(x,y),\cdots,f_m(x,y))$$Suppose there is a point $$(\eta,\xi)$$ such that $$f(\eta,\xi)=\xi$$. Without loss of generality we can assume that $$(\eta,\xi)=0$$. We also assume that $$\partial _{y^j}f_i=0$$, for all $$i,j$$, then we can use the formula that for every $$y,z\in\R^m$$ there is $$\xi\in\R^m$$ in the segment $$[y,z]=\{lx+(1-l)y:l\in[0,1]\}$$$$f_i(x,y)-f_i(x,z)=\sum_j\frac{\partial f_i(x,\xi)}{\partial y^j}(y_j-z_j)$$to prove that given a constant $$00$$ such that in the on $$U\times V$$ $$\|f(x,y)-f(x,z)\|<K\|y-z\|$$where $$U=B(0,a)$$ and $$V=B(0,b)$$. In particular, we can also choose $$a$$ so that on $$U\times V$$, also the following equation holds $$Kb+\|f(x,0)\|<b$$Now, consider the space $$\mathbf F$$ of continuous functions $$U\rightarrow V$$ with the sup norm (since $$V$$ is bounded, then the sup norm is actually well defined), then we can define $$T:\mathbf F \rightarrow\mathbf F$$ $$\zeta\longmapsto [T(\zeta)](x)=f(x,\zeta(x))$$To prove that $$T(\zeta)\in\mathbf F$$, we need to show that $$[T(f)](x)\in V$$, which is a consequence of $$\|f(x,\zeta(x))\|\leqslant \|f(x,0)-f(x,\zeta(x))\|+\|f(x,0)\|$$ and the definition of $$a,b$$. Finally, it easy to prove that $$T$$ is a contraction, so by Banach Contraction Principle there is a unique $$\zeta\in\mathbf F$$ such that $$\zeta(x)=[T(\zeta)](x)=f(x,\zeta(x))$$
 * Implicit Function Theorem (2). Consider a $$C^1$$-differentiable map $$f:\R^n\times\R^m\rightarrow \R^m$$. Suppose

The Fundamental Group

 * Path and homotopy. Given a space $$X$$ and a point $$x\in X$$, we denote with $$\pi_1(X,x)$$ the fundamental group of the space $$X$$. If the space is arc-connected, we denote the fundamental group as $$\pi_1(X)$$. We also denote as $$\pi_0(X,x)$$ the arc-connected component of $$x$$.

Homology groups
(X,U)\longrightarrow 0$$this exact sequence translates into the following exact sequence of homology groups (using the notion of connecting homomorphism) $$H_\bullet(U)\longrightarrow H_\bullet (X)\longrightarrow H_\bullet(X,U)\longrightarrow{} H_{\bullet-1}(U)$$where $$H_\bullet(X,U):=H_\bullet(\mathit\Delta_\bullet(X,U))$$ is called the relative homology of the pair $$(X,U)$$. \otimes G'\longrightarrow \mathit\Delta_\bullet(X)\otimes G\longrightarrow 0$$which translates into a sequence of homology groups $$H_\bullet(X;G)\longrightarrow H_\bullet(X;G')\longrightarrow H_\bullet(X;G)\longrightarrow H_{\bullet-1}(X;G)$$where $$H_\bullet(X;G):=H_\bullet(\mathit\Delta_\bullet(X)\otimes G)$$ is the homology groups with coefficients in $$G$$. In this case, the connecting homomorphism $$H_\bullet(X;G'')\longrightarrow H_{\bullet-1}(X;G)$$ is called Bockstein homomorphism. \longrightarrow^{j_*} H_\bullet(X,A)\longrightarrow^{\partial}H_{\bullet-1}(A)$$is exact. \longrightarrow^{j_*} H_\bullet(X,A)\longrightarrow^{\partial}H_{\bullet-1}(A)$$Using naturality of $$\partial $$ and the fact that both $$A,X \longrightarrow *$$ factor through $$i,j$$ respectively, so we have the induced sequence$$H_{\bullet+1}(X,A)\longrightarrow^{\partial}\tilde H_\bullet(A)\longrightarrow^{i_*}\tilde H_\bullet(X) \longrightarrow^{j_*} H_\bullet(X,A)\longrightarrow^{\partial}\tilde H_{\bullet-1}(A)$$it's easy to prove using definitions that this sequence is exact, and we call this the reduced version of the original exact sequence. \longrightarrow^{j_*} H_\bullet(\mathbf S^n,\mathbf D^n_+)\longrightarrow^{\partial}\tilde H_{\bullet-1}(\mathbf D^n_+)$$since $$i_*=\partial=0$$, we conclude that the map $$j_*:\tilde H_\bullet(\mathbf S^n)\rightarrow H_\bullet(\mathbf S^n,\mathbf D^n_+)$$ is an isomorphism. On the other hand, given $$U$$ a small enough open neighborhood of $$(1,0,\cdots,0)\in\mathbf S^n$$, we have the following sequence of isomorphisms $$H_\bullet(\mathbf S^n,\mathbf D^n_+)\rightarrow^{k_*} H_\bullet(\mathbf S^n-U,\mathbf D^n_+-U)\simeq H_\bullet(\mathbf D^n_-,\mathbf S^{n-1})$$the first induced by excision of $$U$$, while the second from the fact that $$\mathbf S^{n}-U$$ is homotopic to $$\mathbf D^n_-$$ and the same homotopy restricts to an homotopy of $$\mathbf D^n_+-U$$ into $$\mathbf S^{n-1}$$. Thus, finally, consider the reduced sequence$$H_{\bullet+1}(\mathbf D^n_-,\mathbf S^{n-1})\longrightarrow^{\partial}\tilde H_\bullet(\mathbf S^{n-1}) \longrightarrow^{i_*}\tilde H_\bullet(\mathbf D^n_-)\longrightarrow^{j_*} H_\bullet(\mathbf D^n_-,\mathbf S^{n-1})\longrightarrow^{\partial}\tilde H_{\bullet-1}(\mathbf S^{n-1})$$again, since $$\mathbf D^n_-$$ is contractable, we get that $$i_*=j_*=0$$ and $$\partial $$ is an isomorphism. Concluding, we have a sequence of isomorphisms as follows $$\tilde H_\bullet (\mathbf S^n)\simeq H_\bullet(\mathbf S^n,\mathbf D^n_+)\simeq H_\bullet(\mathbf D^n_+,\mathbf S^{n-1}) \simeq\tilde H_{\bullet-1} (\mathbf S^{n-1})$$Therefore, it all comes down to calculating $$\tilde H_\bullet(\mathbf S^0)$$, which is equal to calculating $$H_\bullet(\mathbf S^0,\mathbf D^0_+)\simeq H_\bullet(*\sqcup*,*)\simeq H_\bullet(*)$$therefore we conclude the following homologies and reduced homologies $$H_k(\mathbf S^{n},\mathbf D^n_+)\simeq\tilde H_k(\mathbf S^n)\simeq H_k(\mathbf D^n_+,\mathbf S^{n-1})\simeq\begin{cases}G & k=n\\0 & k\neq n \end{cases}$$Now, suppose $$G$$ is non-zero and that the sphere $$\mathbf S^{n-1}$$ is a retract of $$\mathbf D^{n}_+$$ (meaning there is a map $$f:\mathbf D^n_+\rightarrow\mathbf S^{n-1}$$ such that the composition $$\mathbf S^{n-1}\hookrightarrow\mathbf D^n_+\rightarrow^f\mathbf S^{n-1}$$ is the identity morphism), then we have that the induced composition $$\tilde H_\bullet(\mathbf S^{n-1})\rightarrow\tilde H_\bullet(\mathbf D^n_+)\rightarrow^{f_*}\tilde H_\bullet(\mathbf S^{n-1})$$is the identity, but $$\tilde H_\bullet(\mathbf D^n_+)=0$$, while $$\tilde H_{n-1}(\mathbf S^{n-1})=G$$, absurd, so $$\mathbf S^{n-1}$$ is not a retraction of $$\mathbf D^n_+$$ (provided there is a homology theory with non zero coefficients group). X$$since the inclusion $$Y \hookrightarrow X\vee Y$$ has a retraction (given by the identity $$Y\rightarrow Y$$ and mapping $$X$$ identically to $$x$$), $$j_*$$ is 1-1. For a similar reason, the map $$\pi_*$$ is onto, so we have the sequence $$0\longrightarrow\tilde H_\bullet Y\longrightarrow^{j_*}\tilde H_\bullet(X\vee Y)\longrightarrow^{\pi_*}\tilde H_\bullet X\longrightarrow 0$$At this point, we have that the sequence splits (the inclusions have the projections as retractions), but we need to prove that the sequence is exact in $$\tilde H_\bullet (X\vee Y)$$. To do so, just notice that, if $$N\ni x$$ is a neighborhood of $$x$$ of which the point is a strong deformation retract, then $$Y\hookrightarrow X\vee Y$$ is a deformation retract of $$N\vee Y$$, so we have that the above sequence is isomorphic to $$\tilde H_\bullet N\vee Y \longrightarrow\tilde H_\bullet X\vee Y \longrightarrow H_\bullet(X\vee Y,N\vee Y)$$using excision to cancel $$K=Y-\{y\}$$ from $$X\vee Y$$, the third group becomes $$H_\bullet(X\vee Y,N\vee Y) \simeq H_\bullet(X\vee Y-K,N\vee Y-K)\simeq H_\bullet(X,N)\simeq H_\bullet(X,x)\simeq\tilde H_\bullet X$$This means that the original sequence is exact and that $$\tilde H_\bullet(X\vee Y)\simeq\tilde H_\bullet(X)\oplus\tilde H_\bullet(Y)$$In particular, take $$k$$ copies of the $$n$$-sphere $$\mathbf S^n_i$$ (with $$n>0$$, so that $$H_0\mathbf S^n=0$$ and the reduced homology is equivalent to the classical homology), let $$X=\mathbf S^n_1\vee\cdots\vee\mathbf S^n_k$$, then using the previous result, the map $$ H_n\mathbf S^n_1\oplus\cdots H_n\mathbf S^n_k\rightarrow H_nX$$induced by the inclusions $$j_k:\mathbf S^n_k\hookrightarrow X$$, is an isomorphism, with inverse the map induced by the projections $$\pi_k:X \twoheadrightarrow\mathbf S_k^n$$.
 * Standard $$n$$-simplex. The standard $$n$$-simplex is the following subspace of $$\R^{n}$$ $$\mathit\Delta_n=\{t\in\R^{n+1}|t_j\geqslant 0\text{ for all }j\text{ and }\sum_jt_j=1\}$$For each $$n$$ and $$j\leqslant n+1$$, we define the map $$\delta^n_j:\mathit\Delta_{n-1}\longrightarrow\mathit\Delta_n$$ as $$(t_1,\cdots,t_{n-1})\longmapsto (t_1,\cdots,t_{j-1},0,t_j,\cdots,t_n)$$ we call this map $$j$$-th boundary map. The image of all boundary maps is the boundary of $$\mathit\Delta_n$$.
 * Singular homology. Given a topological space $$X$$, we call a $$n$$-simplex in $$X$$ a continuous map $$\mathit\Delta_n \longrightarrow X$$. Now, let $$\mathit\Delta_n(X) =\text{free abelian group generated by }n\text{-simplexes}$$We also define the following map $$d_n:\mathit\Delta_n(X)\longrightarrow\mathit\Delta_{n-1}(X)$$, called boundary operator as the linear extension of$$d_n\sigma=(\sigma\delta_0)+\cdots+(-1)^{n+1}(\sigma\delta_{n+1})$$for a $$n$$-simplex $$\sigma$$. The previous maps satisfy $$d_{n-1}d_n=0$$ (by convention $$d_0=0$$), so let $$\mathbf{ker}d_n$$ be the group of $$n$$-cycles and $$\mathbf{im}d_{n+1}$$ the group of $$n$$-boundaries, then $$\mathbf{im}d_{n+1}\subseteq\mathbf{ker}d_n$$ and we can define the following$$H_n(X)=\frac{\mathbf{ker}d_n}{\mathbf{im}d_{n+1}}$$called the $$n$$-th singular homology group.
 * Relative homology. Consider a subspace $$U\hookrightarrow X$$, the inclusion induces a map $$\mathit\Delta_\bullet(U)\hookrightarrow \mathit\Delta_\bullet(X)$$, thus we can define the following $$\mathit\Delta_\bullet(X,U):=\mathit\Delta_\bullet(X)/\mathit\Delta_\bullet(U)$$and the corresponding exact sequence$$0\longrightarrow\mathit\Delta_\bullet(U)\longrightarrow\mathit\Delta_\bullet(X)\longrightarrow\mathit\Delta_\bullet
 * Homology with coefficients. Consider a short exact sequence of abelian groups $$0 \longrightarrow G\longrightarrow G'\longrightarrow G''\longrightarrow 0$$since $$\mathit\Delta_\bullet(X)$$ is a free abelian group, then tensor with $$\mathit\Delta_\bullet(X)$$ preserves exactness of short sequences, therefore we have the following exact sequence $$0\longrightarrow \mathit\Delta_\bullet(X)\otimes G\longrightarrow \mathit\Delta_\bullet(X)
 * Axioms for homology. Let $$S(\mathbf{Grp})$$ be the category of functors $$\mathbb N\longrightarrow\mathbf{Grp}$$ and natural transformations between them. Also, let $$\mathfrak{htop}$$ be the category of pairs $$(X,A)$$, where $$A$$ is a subspace of $$X$$ . A homology theory is a functor $$H:\mathfrak{htop}\longrightarrow S(\mathbf{Grp})$$together with a natural transformation $$\partial=\partial_{(X,A)}:H_{+1}(X,A)\longrightarrow H(A)$$, called connecting morphism, where $$H_{}(A):= H_{}(A,\O),\qquad H_{+1}(X,A):n\longmapsto H_{n+1}(X,A) $$($$H_{}(-)$$ is defined as $$H$$ on morphisms, since a morphism $$(X,A)\longrightarrow (Y,B)$$ is also a morphism $$(A,\O)\longrightarrow (B,\O)$$). The pair $$(H,\partial)$$ must satisfy the following axioms, called Eilenberg-Steenrod-Milnor axioms (ESM axioms):
 * (Homotopy axiom). Given two maps $$f,g:(X,A)\longrightarrow(Y,B)$$, if there is an homotopy $$f\simeq g$$, then $$f_*=g_*:H_\bullet(X,A)\longrightarrow H_\bullet(Y,B)$$That is, homotopic maps induce the same morphism between homologies.
 * (Exactness axioms). For every pair $$(X,A)$$, consider the inclusions $$i:(A,\O)\hookrightarrow (X,\O)$$ and $$j:(X,\O)\hookrightarrow (X,A)$$, then the following sequence $$H_{\bullet+1}(X,A)\longrightarrow^{\partial}H_\bullet(A)\longrightarrow^{i_*}H_\bullet(X)
 * (Excision axioms). Given a pair $$(X,A)$$ and an open subspace $$U\hookrightarrow X$$ such that $$\text{c}\ell U\subseteq\text{int}A$$, then we require that the inclusion $$k:(X-U,A-U)\hookrightarrow (X,A)$$ induce a map$$k_*:H_\bullet(X-U,A-U)\longrightarrow H_\bullet(X,A)$$  which is an isomorphism.
 * (Dimension axiom). Let $$*$$ be the one point space, then $$H_n(*)=0$$for all $$n>0$$. The group $$H_0(*)=G$$ is called coefficient group of the theory $$(H,\partial)$$.
 * (Additivity axioms). Given a family of spaces $$\{X_\alpha\}_{\alpha}$$, let $X=\bigsqcup_\alpha X_\alpha$, then we have canonical immersions $$j_\alpha:X_\alpha\hookrightarrow X$$ and we require that the corresponding map $$\sum_\alpha (j_\alpha)_*:\sum_\alpha H_\bullet(X_\alpha)\longrightarrow H_\bullet(X)$$is an isomorphism.
 * Reduced homology. Consider a space $$X$$ and the unique morphism $$p:X\rightarrow*$$, which is a split epimorphism, which section any function $$*\rightarrow X$$ (which corresponds to a unique point in $$X$$), thus the induced morphism $$p_*:H(X)\longrightarrow H(\mathbf p)$$is a split epimorphism. We define $$\tilde H(X)=\mathbf{ker}p_*$$ as the reduced homology of $$X$$, which, by definition, induces the following exact sequence $$\tilde H(X)\hookrightarrow H(X)\twoheadrightarrow^{p_*}H(*)$$Now, consider $$H_\bullet(\O)$$, since $$\O=\O\sqcup \O$$, the canonical isomorphism $$H_\bullet(\O)\oplus H_\bullet(\O)\rightarrow H_\bullet(\O)$$ defined as $$(z,x)\mapsto z+x$$ is an isomorphism, which means that $$H_\bullet(\O)=0$$ (since if there was a non zero element $$x$$, the points $$(0,x),(x,0)$$ would be different but with the same image), so for every space $$X$$, $$H_\bullet(X,X)\simeq H_\bullet(\O,\O)=0$$This means that the map $$H_\bullet(X,A)\longrightarrow H_\bullet(*,*)$$induced by the unique $$X \rightarrow *$$, is a zero homomorphism. Now, Take the induced exact sequence by the inclusion $$i:A\hookrightarrow X$$ $$H_{\bullet+1}(X,A)\longrightarrow^{\partial}H_\bullet(A)\longrightarrow^{i_*}H_\bullet(X)
 * Homology of the spheres. Consider $$\mathbf S^{n}=\{x\in\mathbb R^{n+1}:\|x\|=1\},\qquad\mathbf D^n_+=\{x\in\mathbf S^n:x_0\geqslant 0\}$$ Notice that we have the following two inclusions $$\mathbf S^{n-1}\hookrightarrow\mathbf D^n_+\hookrightarrow\mathbf S^n$$ and that $$\mathbf D^n_+$$ is homeomorphic to the $$n$$-ball $$\mathbf B^n=\{x\in\R^n:\|x\|\leqslant 1\}$$which is a contractable space, therefore $$\tilde H_\bullet(\mathbf D^n_+)\simeq\tilde H_\bullet(\mathbf B^n)=0$$. So, first consider the reduced chain $$H_{\bullet+1}(\mathbf S^n,\mathbf D^n_+)\longrightarrow^{\partial}\tilde H_\bullet(\mathbf D^n_+)\longrightarrow^{i_*}\tilde H_\bullet(\mathbf S^n)
 * Degree morphism. Consider an homology theory $$H$$ with $$G=\mathbb Z$$, then every map $$f:\mathbf S^n\rightarrow\mathbf S^n$$ induces a morphism $$\Z\simeq\tilde H_n(\mathbf S^n)\rightarrow^{f_*}\tilde H_n(\mathbf S^n)\simeq\Z$$which is identified by the unique image $$\delta f=f_*(1)$$ (in particular, $$f_*(n)=(\delta f)n$$). We thus have a morphism $$\delta:C(\mathbf S^n,\mathbf S^n)\longrightarrow\Z$$which preserves composition and identity (so a morphism of monoids). In particular, define $$\iota:\mathbf S^n\rightarrow\mathbf S^n$$ as $$(\vec x,x_n)\longmapsto(\vec x,-x_n)$$this map is easy to see that it commutes with the isomorphism $$\varphi:\tilde H_\bullet\mathbf S^n\longrightarrow \tilde H_{\bullet-1}\mathbf S^{n-1}$$given above (if we take $$U$$ so that it is invariant under $$\iota$$). Then, suppose we proved $$\delta \iota=-1$$ for $$\iota:\mathbf S^{n-1}\rightarrow\mathbf S^{n-1}$$, then $$\varphi \iota_*(x)=\iota_*(\varphi(x))=(\delta \iota)\varphi(x)=-\varphi(x)=\varphi(-x)$$so $$\iota_*(x)=-x$$ and $$\delta \iota=-1$$. To conclude that $$\delta \iota=-1$$, we simply need to prove it for $$\iota:\mathbf S^0\longrightarrow\mathbf S^0$$. Notice that $$\mathbf S^0=*\sqcup*$$ and under the isomorphism $$H_\bullet(*)\oplus H_\bullet(*)\rightarrow H_\bullet\mathbf S^0$$$$\iota_*$$ corresponds to $$(x,y)\longmapsto(y,x)$$, while the induced map $$H_\bullet\mathbf S^0\rightarrow H_\bullet*$$ corresponds to $$(x,y)\longmapsto x+y$$, so that $$\tilde H_\bullet\mathbf S^0\simeq\{(x,-x):x\in H_\bullet (*)\},\qquad \iota_*(x,-x)=(-x,x)=-(x,-x)$$so $$\delta \iota=-1$$, as we wanted to prove: The degree of the map $$\mathbf S^n\rightarrow\mathbf S^n$$ defined as $$(\vec x,x_n)\longmapsto(\vec x,-x_n)$$ is $$-1$$ for all $$n$$. In particular, notice that the antipodal map $$\vec x\longmapsto -\vec x$$ in $$\mathbf S^n$$ is just the product of $$n+1$$ maps like $$f$$ above, so $$\delta(\text{antipodal})=(-1)^{n+1}$$this implies that, for $$n$$ even, the antipodal map cannot be homotopic to the identity. In particular, suppose there is a map $$f:\mathbf S^{2n}\rightarrow\mathbf S^{2n}$$ such that $$f(x)\neq\pm x$$ for every $$x$$, then are well defined the two maps$$H(x,t)=\frac{(1-t)x+tf(x)}{\|(1-t)x+tf(x)\|},\qquad K(x,t)=\frac{(1-t)x-tf(x)}{\|(1-t)x-tf(x)\|}$$since $$\|x\pm tf(x)\|$$ is zero if and only if $$(1-t)x=\pm tf(x)$$, which implies (since both $$f(x),x$$ are unit vectors, so $$1-t=t$$) that $$f(x)=\pm x$$, which cannot be. So we have an homotopy $$H:1\rightarrow f$$ and $$K:\text{antipodal}\rightarrow f$$, thus there is an homotopy $$1\rightarrow\text{antipodal}$$contary to the last result, so For every map $$f:\mathbf S^{n}\rightarrow\mathbf S^{n}$$, with $$n$$ even, there is a point $$x$$ such that $$f(x)=\pm x$$. Finally, identify the tangent space $$T_x\mathbf S^n$$ with the plane $$\langle x\rangle^\perp$$ (that is, $$T\mathbf S^n$$ is the collection of points $$x+v$$, with $$x\in\mathbf S^n$$ and $$v\perp x$$). Now, suppose $$v:\mathbf S^n \rightarrow T\mathbf S^n $$ is a non-zero vector field, which corresponds to a map $$v:\mathbf S^n\rightarrow\mathbb R^{n+1}$$ sending a point $$x$$ to a vector $$v_x\perp x$$. Consider then the map $$f:\mathbf S^n\ni x\longmapsto \frac{v_x}{\|v_x\|}\in\mathbf S^n$$then there is $$x$$ such that $$x\perp v_x/\|v_x\|=\pm x$$, absurd, so For even $$n$$, every vector field $$v$$ on $$\mathbf S^n$$ has a zero at some point. Consider, instead, a sphere $$\mathbf S^{2n-1}\hookrightarrow\R^{2n}$$, this actually has a nowhere zero vector field, namely, $$(x_0,\cdots,x_{n-1},x_{n},\cdots,x_{2n-1})\longmapsto(-x_n,\cdots,-x_{2n-1},x_0,\cdots,x_{n-1})$$
 * Wedge sum. Consider two pointed spaces $$(X,x)$$ and $$(Y,y)$$, the we define the wedge sum, written $$X\vee Y$$, as the space $$X\sqcup Y$$ modulo the identification $$x=y$$. The sequence of maps $$Y\hookrightarrow^j X\vee Y \twoheadrightarrow^\pi X$$ induces a sequence $$\tilde H_\bullet Y\longrightarrow^{j_*}\tilde H_\bullet(X\vee Y)\longrightarrow^{\pi_*}\tilde H_\bullet

CW Complexes
Now, (a) clearly is true for $$K^{(0)}$$, so (c) is true as well. If (c) is true for $$K^{(n-1)}$$, then (b) is true, so $$f_\sigma(\mathbf S^{n-1})\subseteq K^{(n-1)}$$ is covered by finitely many open cells, so $$f_{\sigma}(\mathbf S^{n-1})\subseteq\text{finitely many open cells}\subseteq\text{finitely many cells}\subseteq \text{finitely many finite subcomplexes}$$so $$f_\sigma(\mathbf S^{n-1})$$ is contained in a finite subcomplex $$K'$$ (union of the finitely many finite subcomplexes containing it), so if we add the cell $$f_\sigma:\mathbf D^{n}\rightarrow K'$$, we have a finite subcomplex containing $$C_\sigma$$, thus (c) is true for $$K^{(n)}$$. So In every CW complex $$K$$, the following three are equivalent and valid $$) covers $$C$$.  We need to prove that $$C$$ has the weak topology defined by $$f_\overline{\sigma}$$, that is, given $$A\hookrightarrow C$$, if $$f^{-1}_ \overline{\sigma}(A)$$ is open for every $$\overline\sigma$$, then $$A$$ is open. Now, consider the $$U\hookrightarrow C$$ such that $$p|_{U}:U\rightarrow p(U)$$ is an homeomorphism, then $$f^{-1}_ \overline{\sigma}(A\cap U)=f^{-1}_\overline{\sigma}(A)\cap f^{-1}_\overline{\sigma}(U)$$and $A=\bigcup_U A\cap U$, so we can assume $$A\hookrightarrow U$$, for some $$U$$. Now, if $$x\in f^{-1}_\sigma(p(A))$$, then $$f_\sigma(x)=p(a)$$ for some $$a\in A$$, so since there is a lift $$f_\overline{\sigma}$$ such that $$f_ \overline{\sigma}(x)=a$$$$f^{-1}_\sigma(p(A))=\bigcup\{f^{-1}_\overline{\sigma}(A)|f_\overline{\sigma}\text{ lift of }f_\sigma\}$$so $$f^{-1}_\sigma(p(A))$$ is the union of open sets, so it is open for every $$\sigma$$. Thus, having $$K$$ the weak topology, $$p(A)$$ is open and $$A$$ itself is open.
 * CW complexes. A CW complex is a topological space $$K$$ constructed as the union of other spaces $$K^{(0)},\cdots,K^{(n)},\cdots$$ defined by induction as follows
 * $$K^{(0)}$$ is a non-empty discrete space (that is, a simple set of points)
 * Once we have $$K^{({n-1})}$$, a (eventually empty) family of continuous maps $$\{f_\sigma:\mathbf S^{n-1}\rightarrow K^{(n-1)}\}_\sigma$$, called attachment maps, then $$K^{(n)}$$ is the following quotient space $$K^{(n-1)}\sqcup(\mathbf D^n\times I)/\sim$$where $$I$$ is the index (with discrete topology) of the family $$f_\sigma$$ and $$\sim$$ is the smallest equivalence that identifies $$x\in K^{(n-1)}$$ with $$(y,\sigma)\in\mathbf D^n\times I$$ iff $$f_\sigma(y)=x$$.
 * Finally, $$K=K^{(0)}\cup\cdots\cup K^{(n)}\cup\cdots$$, with the weak topology induced by the different maps $$f_\sigma:\mathbf D^{n}\rightarrow K$$. We call the subset $$U_\sigma=f_\sigma(\mathbf D^n-\mathbf S^{n-1})$$ open cell (a single $$x\in K^{(0)}$$ is an open cell), while $$C_\sigma=f_\sigma(\mathbf D^n)$$ is a closed cell. Notice that every point in $$K$$ belongs to an open cell.
 * Properties of CW complexes. Consider a CW complex $$K$$ and suppose that $$A\hookrightarrow K$$ is closed and discrete if it has no two points in the same open cell (a).   Consider a compact $$C\hookrightarrow K$$ and select the set $$D\hookrightarrow C$$ by taking a point from every open cell intersecting $$C$$, then $$D$$ is finite (thus compact, being a closed subspace of a compact one) and discrete, thus $$D$$ is finite and Every compact subset is covered by finitely many open cells (b).     Now, consider a cell $$C_\sigma=f_\sigma(\mathbf D^n)$$. This form a closed $$n$$-cell which is covered by $$U_\sigma$$ and $$K^{(n-1)}$$ and $$C_\sigma\cap K^{(n-1)}=f_\sigma(\mathbf S^{n-1})$$which is compact, thus can be covered by finitely many open cells in $$K^{(n-1)}$$, so open $$n-1$$-cells. Continuing, we arrive at $$0$$-cells, thus we formed a subcomplex $$K'$$ with finitely many attachment maps (thus a finite subcomplex) containing $$C_\sigma$$. So, Every cell $$C_\sigma$$ is contained in a finite subcomplex (c).    Finally, consider $$A$$ intersecting every open cell in one element only. Given any cell $$C_\sigma$$, this is contained in a finite subcomplex $$K'$$, so $$A\cap C_\sigma\subseteq A\cap K'$$But $$A$$ intersect every open cell of $$K'$$ only once and, since $$K'$$ is a finite subcomplex, there are only a finite number of them, so $$A\cap C_\sigma$$ is finite and thus closed for every $$\sigma$$, thus $$A$$ is closed (by weak topology). Now, given any $$x\in A$$, the complement $$A-\{x\}$$ satisfy the same condition as $$A$$, thus $$A-\{x\}$$ is closed and $$\{x\}$$ is open in $$A$$. If a set $$A$$ intersects every open cell in one point, then it is closed and discrete.
 * If a set $$A$$ intersects every open cell in one point, then it is closed and discrete
 * Every compact subset is covered by finitely many open cells
 * Every cell $$C_\sigma$$ is contained in a finite subcomplex
 * Covering of a CW complex. Consider a CW complex $$K$$ and a covering $$p:C\rightarrow K$$. Since $$\mathbf D^n$$ is path connected and simply connected, every map $$f_\sigma:\mathbf D^n\rightarrow K$$ has a family of lifts $$f_{\overline{\sigma}} :\mathbf D^n\rightarrow C$$ such that the images of every lift $$f_\overline{\sigma}$$ (for every attachment map $$f_\sigma

Singular Homology

 * Chain homotopy. Given two chains $$C,D$$, two morphisms $$f,g:C \rightarrow D$$ are chain homotopic, written $$\phi\simeq\psi$$, if there is a sequence of homomorphisms $$\delta=\delta_j:C_j\rightarrow D_{j+1}$$ such that $$f-g=d\delta+\delta d$$. In particular, two chain homotopic morphisms induce that same morphism $$H_\bullet C\rightarrow H_\bullet D$$.