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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf  Topologia das Variedades \\[15mm]
\small Cohomology, lecture 5: Mayer-Vietoris exact sequence}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small\bf Misha Verbitsky  
\\[2mm]
May 10, 2018\\[4mm]

{\tiny\bf IMPA, Tuesdays and Thursdays, 10:30, Sala 224}
}
\end{center}

\newpage


{\bf \blue Pullback of a differential form (reminder)}

\definition
Let  $M \stackrel \phi\arrow N$ be a morphism of smooth manifolds,
and $\alpha \in \Lambda^i N$ be a differential form.
Consider an $i$-form $\phi^* \alpha$ taking value
\[ \alpha\restrict{\phi(m)}(D_\phi(x_1), ... D_\phi(x_i))\] on
$x_1, ..., x_i\in T_m M$. It
is called {\bf\blue the pullback of $\alpha$}. 
If $M\stackrel \phi \arrow N$ is a closed embedding,
the form $\phi^* \alpha$ is called
{\bf \blue the restriction} of $\alpha$ to 
$M \hookrightarrow N$.

\lemma {\bf \blue (*)}
Let $\Psi_1, \Psi_2:\; \Lambda^* N \arrow \Lambda^* M$
be two maps which satisfy graded Leibnitz identity, supercommutes with
de Rham differential, and satisfy \\ $\Psi_1\restrict{C^\infty M}=
\Psi_2\restrict{C^\infty M}$. {\bf \red Then $\Psi_1 = \Psi_2$.}

\proof
The algebra $\Lambda^* M$ is generated multiplicatively
by $C^\infty M$ and $d(C^\infty M)$; restrictions of $\Psi_i$
to these two spaces are equal. \endproof


\claim
{\bf \red Pullback commutes with the de Rham differential.}

\proof
Let $d_1, d_2:\; \Lambda^* N \arrow \Lambda^{*+1} M$
be the maps $d_1 = \phi^* \circ d$ and $d_2= d\circ \phi^*$.
{\bf \purple These maps satisfy the Leibnitz identity, and they are equal
on $C^\infty M$.} The super-commutator $\delta:= \{d_i, d\}$
is equal to $d \circ\phi^* \circ d$, it commutes 
with $d$, and equal 0 on functions. By Lemma (*), $\delta=0$.
Then $d_i$ supercommutes with $d$. Applying Lemma (*) again,
we obtain that $d_1=d_2$.
\endproof


\newpage

{\bf \blue Lie derivative (reminder)}

\definition
Let $B$ be a smooth manifold, and $v\in TM$ a vector field.
An endomorphism $\Lie_v:\; \Lambda^* M \arrow \Lambda^* M$,
preserving the grading is called
{\bf\blue a Lie derivative along $v$} if it satisfies the 
following conditions.\\
\phantom{XX} (1) On functions $\Lie_v$ is 
equal to a derivative along $v$.  (2) $[\Lie_v, d]=0$.\\
 \phantom{XX} (3) $\Lie_v$ is a derivation of the de Rham algebra.

\remark
The algebra $\Lambda^*(M)$ is generated 
by $C^\infty M=\Lambda^0(M)$ 
and $d(C^\infty M)$. The restriction
$\Lie_v\restrict{C^\infty M}$ is determined by the first axiom.
On $d(C^\infty M)$ is also determined
because $\Lie_v(df) =d(\Lie_v f)$. {\bf \red Therefore,
$\Lie_v$ is uniquely defined by these axioms.}

\theorem
{\bf \blue (Cartan's formula)}
Let $i_v$ be a convolution with a vector 
field, $i_v(\eta) = \eta(v, \cdot, \cdot, ..., \cdot)$
{\bf \red Then $\{d, i_v\}$ is equal to the Lie derivative along $v$.}


\newpage

{\bf \blue Flow of diffeomorphisms (reminder)}


\definition
Let $f:\; M \times [a,b]\arrow M$ be a smooth map
such that for all $t\in [a,b]$ the restriction
$f_t:= f\restrict{M\times\{t\}}:\; M \arrow M$
is a diffeomorphism. Then $f$ is called {\bf\blue a flow
of diffeomorphisms}. 

\claim
Let $V_t$ be a flow of diffeomorphisms, $f\in C^\infty M$,
and $V^*_t(f)(x):= f(V_t(x))$. Consider the map
$\frac d{dt}V_t\restrict{t=c}:\; C^\infty M \arrow C^\infty M$,
with $\frac d{dt}V_t\restrict{t=c}(f)=
(V_c^{-1})^*\frac{dV_t}{dt}\restrict{t=c}f$. 
{\bf \red Then
$f\arrow (V_t^{-1})^*\frac d{dt}V_t^*f$ is a derivation} 
(that is, a vector field).

\theorem
Let $M$ be a compact manifold, and $X_t\in TM$ a family
of vector fields smoothly depending on $t\in [0,a]$.
{\bf \red Then there exists a unique diffeomorphism flow $V_t$,  $t\in [0,a]$,
such that $V_0=\Id$ and $\frac d{dt}V_t^*=X_t$.}


\newpage

{\bf \blue Lie derivative and a flow of diffeomorphisms (reminder)}

\definition
Let $v_t$ be a vector field on $M$, smoothly depending on 
the ``time parameter'' $t\in[a,b]$, and
$V:\;M\times[a,b]\arrow M$ a flow of diffeomorphisms which satisfies
$\frac d{dt}V_t=v_t$ for each $t\in[a,b]$, and $V_0=\Id$.
Then $V_t$ is called {\bf\blue an exponent of $v_t$}.

\claim 
Exponent of a vector field is unique; it exists when $M$
is compact. This statement is called {\bf \blue ``Picard-Lindel\"of theorem''}
or {\bf \blue ``uniqueness and existence of solutions of ordinary differential
equations''}. 

\proposition
Let $v_t$ be a time-dependent vector field, $t\in[a,b]$,
and $V_t$ its exponent.
For any $\alpha \in \Lambda^*M$, consider $V_t^*\alpha$
as a $\Lambda^*M$-valued function of $t$.
{\bf \red Then $\Lie_{v_t}(\alpha) = (V_t^{-1})^*\frac{d}{dt}(V_t^*\alpha)$.}

\newpage

{\bf \blue Lie derivative and cohomology (reminder)}

\claim
Let $v$ be a vector field, and $\Lie_v:\; \Lambda^* M\arrow \Lambda^* M$
be the corresponding Lie derivative. Then {\bf \red $\Lie_v$
commutes with the de Rham differential, and acts
trivially on the de Rham cohomology.}

\proof 
$\Lie_v=i_v d+di_v$ maps closed forms to exact. \endproof

\corollary
Let $V_t$, $t\in[a, b]$ be a flow of diffeomorphisms on a manifold $M$.
{\bf \red 
Then the pullback map 
$V_b^*$ acts on cohomology the same way as $V_a^*$.}

\proof
Since $(V_t^{-1})^*\frac {d V^*_t}{dt}(\eta)= \Lie_{X_t}(\eta)$, 
this map it acts trivially on cohomology. Then 
$V_b^*-V_a^*(\eta)=\int_{a}^b V^*_t\Lie_{X_t}(\eta)$ is exact for any closed $\eta$.
Therefore, $V_b^*(\eta)-V_a^*(\eta)$ is exact.
\endproof

\newpage

{\bf \blue Extension of vector fields}

\exercise
Let $M \subset M_1$ be a smooth closed submanifold and
$v \in TM_1 \restrict M$ a tangent to $M_1$ vector field
defined on $M$. {\bf \purple Prove that $v$ can be extended to a vector field
on $M_1$.} Moreover, this extension can be chosen smoothly for any
smooth family of vector fields $v$.

\newpage

{\bf \blue Homotopy equivalence of cohomology}

\theorem
Let $F_t:\; X \arrow Y$ be morphisms (smooth maps) of compact 
manifolds, smoothly depending on $t\in [a, b]$. Consider the
corresponding maps on cohomology: $F_t^*:\; H^*(Y)\arrow H^*(X)$.
{\bf \red Then $F^*_a= F^*_b$.}

\pstep
Consider the graph $\Gamma_t \subset X\times Y$ of $F_t$.
Then $\Gamma_t$ is a smooth submanifold in $X\times Y$,
and $v_x:=\frac d{dt}F_t(x)\in T_{F_t(x)} Y$ gives a 
tangent vector $(0, v_x)\in T_{x, F_t(x)}  X\times Y= T_x Y \times T_{F_t(x)}X$.
Taking all $v_x$ for all points $(x, F_t(x))\in \Gamma_t$,
we obtain a vector field $\beta_t \in T(X\times Y)\restrict{\Gamma_t}$.

{\bf \green Step 2:} Using the previous exercise, we extend
$\beta_t$ to a vector field on $X\times Y$.
The corresponding flow of diffeomorphisms $V_t:\; X\times Y \arrow X\times Y$
maps $(x, F_a(x))$ to $(x, F_t(x))$. 

{\bf \green Step 3:} Since the flows of diffeomorphisms
induce constant maps on cohomology,  the natural restriction maps 
$H^*(X\times Y)\arrow H^*(\Gamma_a)= H^*(X)$ and
$H^*(X\times Y)\arrow H^*(\Gamma_b)= H^*(X)$ are equal.
However, the pullback map $F^*_t:\; H^*(Y)\arrow H^*(X)$
can be obtained as a composition of the pulback map
$H^*(Y) \arrow H^*(X\times Y)$ associated with the projection,
and the restriction $H^*(X\times Y)\arrow H^*(\Gamma_t)= H^*(X)$.
Therefore $F^*_a=F^*_b$, and the de Rham cohomology are
homotopy invariant.
\endproof


\newpage

\vfill

\begin{center}{\bf \red \large Tuesday lecture\\ is moved to
May 14, Monday, 10:30, room 228}
\end{center}

\hfill



\newpage

{\bf \blue Brouwer fixed point theorem: a preparatory lemma}

\lemma
Let $B\subset \R^n$ be a closed ball, 
and $\Psi:\; B \arrow \6 B$ be a map to its boundary $\6 B$
such that $\Psi\restrict{\6 B}= \Id_{\6 B}$. {\bf \red Then 
$\Psi$ cannot be smooth.}

\proof Suppose that $\Psi$ is smooth. 
Consider the composition $j \circ \Psi$ of embedding $j:\; \6 B \arrow B$ 
and $\Psi$. Since $H^{n-1}(S^{n-1})= \R$ and $j \circ \Psi=\Id_{\6 B}$,
the map $\Psi^*\circ j^*$ acts as identity on $H^{n-1}(S^{n-1})$.
This is impossible because {\bf \purple by Poincar\'e lemma
$H^{n-1}(B)=0$, hence $j^*:\; H^{n-1}(B)\arrow H^{n-1}(S^{n-1})$
is zero, and composition of $\Psi^*$ and $j^*$ vanishes
on $H^{n-1}(S^{n-1})$.}
\endproof

\newpage

{\bf \blue Brouwer fixed point theorem}


\theorem {\bf \blue (Brouwer fixed point theorem)}\\
Let $B\subset \R^n$ be a closed ball, 
and $\phi:\; B \arrow B$ be a continuous map.
{\bf \red Then $B$ has at least one fixed point.}

\pstep Suppose that $\phi$ has no fixed points.
Since $B$ is compact, {\bf \purple there exists $\epsilon >0$ such that
$d(x, \phi(x))>2 \epsilon$ for each $x\in B$.}

{\bf \green Step 2:} By Stone-Weierstrass, any
continuous map can be $\epsilon$-approximated by a smooth map
$\phi_1$ such that  $d(x, \phi_1(x))\geq 
d(x, \phi(x))- d(\phi(x), \phi_1(x)) > \epsilon$.
This implies that $\phi_1$ has no fixed points,
and {\bf \purple we may assume that $\phi$ is smooth.}

{\bf \green Step 3:} Let $l_x\subset \R^n$ be an oriented line 
connecting $x$ to $\phi(x)$. Then $l_x$ intersects the boundary
$\6 B$ in two points, distinguished by its orientation.
Let $\Psi(x)$ be the first point of intersection
of $l_x$ and $\6B$. Clearly, {\bf \purple the map $x \arrow \Psi(x)$
is smooth and $\Psi(x)=x$ for all $x\in \6 B$.}
This is impossible by the previous lemma.
\endproof


\newpage

{\bf \blue Long exact sequence}

\definition
{\bf\blue A complex} is a sequence of vector spaces
and homomorphisms
$... \stackrel d\arrow C^{i-1} \stackrel d\arrow
C^i\stackrel d\arrow C^{i+1} \stackrel d\arrow...$
satisfying $d^2=0$.  {\bf \blue
Homomorphism} $(C^*, d)\arrow (C_1^*, d)$ of complexes 
is a sequence of homomorphism $C^i \arrow C_1^i$
commuting with the differentials.

\definition
An element $c\in C^i$ is called {\bf\blue closed}
if $c\in \ker d$ and {\bf \blue exact} if $c\in \im d$.
{\bf \blue Cohomology} of a complex
is a quotient $\frac{\ker d}{\im d}$.
One denotes the $i$-th group of cohomology 
of a complex by $H^i(C^*)$

\remark A homomorphism of complexes induces a natural
homomorphism of cohomology groups.

\definition
{\bf \blue Short exact sequence of complexes}\\
$0 \arrow A^* \arrow B^* \arrow C^* \arrow 0$
is a sequence of morphisms of complexes
$A^* \stackrel x \arrow B^* \stackrel y\arrow C^*$
such that $x:\; A^i \arrow B^i$ is injective,
$y:\; B^i \arrow C^i$ is surjective (for all $i$),
and $\ker y= \im x$.

\theorem
Let  $0 \arrow A^* \arrow B^* \arrow C^* \arrow 0$
be an exact sequence of complexes. Then 
there exists a {\bf \blue long exact sequence of cohomology}
\[... \arrow H^{i-1}(C^*) \arrow H^i(A^*) \arrow  H^i(B^*) 
\arrow H^{i}(C^*) \arrow H^{i+1}(A^*) \arrow ...
\]

\newpage

{\bf \blue Differential forms on closed subsets}

\definition
Let $M$ be a manifold, and $Z\subset M$
a closed subset. Let $\alpha, \beta$
be two differential forms defined in 
open sets $U_\alpha$ and $U_\beta$ containing $Z$.
We say that $\alpha$ and $\beta$ are equivalent
if $\alpha=\beta$ on $U_\alpha\cap U_\beta$.
The space of equivalence classes is denoted
$\Lambda^*(Z)$ and called {\bf \blue the space of
differential forms on $Z$} or {\bf \blue
the space of germs of differentials forms in $Z$}.

\remark
{\bf \purple If $Z$ is a manifold with boundary, 
$\Lambda^*(Z)$ is the space of differential
forms on $Z$,} by definition of differential
forms on manifolds with boundary.

\exercise
Using partition of unity, {\bf \red prove that
the natural restriction map $\Lambda^*(M)\arrow \Lambda^*(Z)$
is surjective for any closed $Z\subset M$.}

\remark
The de Rham differential is well defined on $\Lambda^*(Z)$,
allowing us to {\bf \purple define the de Rham cohomology of $Z$ as usual.}
Poincar\'e lemma holds in this situation, too.

\claim
Let $X\subset \R^n$ be a closed starlike subset.
{\bf \red Then $H^i(X)=0$ for all $i>0$.}

\proof Same as for open $X$. \endproof

\newpage

{\bf \blue Mayer-Vietoris long exact sequence}

\claim
Let $X, Y\subset M$ be closed subsets. Then 
{\bf \red the restriction maps define an exact sequence of complexes:}
\[ 
0 \arrow \Lambda^*(X\cup Y)\stackrel \phi \arrow \Lambda^*(X)
\oplus  \Lambda^*(Y) \stackrel \psi \arrow \Lambda^*(X\cap Y)\arrow 0. \ \ (*)
\]
Here $\phi$ is restriction to both components, and
$\psi$ is restriction $|_{X\cap Y}$
on the first component and $-|_{X\cap Y}$ on the second component.

\proof
The map $\phi$ is clearly injective;  $\psi$
is surjective because all forms in $\Lambda^*(X\cap Y)$
can be smoothly extended to $M$. Now, $\ker \psi$ 
is pairs $\alpha \in \Lambda^*(X),  \beta \in \Lambda^*(Y)$
such that $\alpha|_{X\cap Y}=- \beta|_{X\cap Y}$,
and this is precisely pairs which agree on an open neighbourhood of
$X\cap Y$, that is, obtained by restriction from some 
$\gamma\in \Lambda^*(X\cup Y)$.
\endproof

\corollary {\bf \blue (Mayer-Vietoris long exact sequence)}
Let $X, Y\subset M$ be closed subsets.
{\bf \red Then there is a long exact sequence associated with (*):}
\begin{multline*}... \arrow H^{i-1}(X)
\oplus H^{i-1}(Y) \arrow H^{i-1}(X\cap Y) \arrow H^i(X\cup Y) \arrow \\ \arrow  H^i(X)
\oplus H^i(Y)  
\arrow H^{i}(X\cap Y) \arrow H^{i+1}(X\cup Y) \arrow ...
\end{multline*}
\endproof

\newpage

{\bf \blue Computation of de Rham cohomology}

\theorem
Let $M$ be a smooth manifold. {\bf \red Then $M$ admits a 
polyhedral structure and a triangulation.}

\proof Was proven earier.
\endproof

\corollary Let $M$ be a compact smooth manifold. {\bf \red Then 
the cohomology of $M$ are finite-dimensional.}

\pstep
Suppose that $M$ is a union of $n$
closed subsets $K_1$, ..., $K_n$ which are starlike in some coordinates, 
and such that any intersection of any family of $K_i$ is also starlike.
Using induction by $n$, we may assume that the union of 
$n-1$ starlike sets with this property has finite cohomology.
Consider the Mayer-Vietoris exact sequence
\[...   \arrow H^{i-1}(X\cap Y) \arrow H^i(X\cup Y) \arrow  H^i(X)
\oplus H^i(Y)  
\arrow ...
\]
where $X=K_1$ and $Y=\bigcup_{i=2}^n K_i$.
Induction assumption gives that $H^i(Y)$ and $H^i(X\cap Y)$
has finite-dimensional cohomology, and $H^i(X)$ is finite-dimensional
by Poincar\'e lemma. Then $H^i(X\cup Y)$ is also finite-dimensional.

{\bf \green Step 2:} For any triangulation or a polyhedral structure,
the simplices or polyhedra and their intersections are starlike,
hence satisfy assumptions of Step 1. \endproof

\newpage

{\bf \blue Singular cohomology}

\definition
Let $M$ be a topological space and $C_i$ {\bf \blue the group of $i$-chains,}
that is, the group freely generated by continuous maps $f:\; \Delta^i \arrow M$,
where $\Delta^i$ is $i$-simplex. {\bf \blue Boundary operator}
maps a $k$-simplex $f$ to a sum $\sum_{i=0}^k (-1)^k f_i$,
where $f_i\in C_{i-1}$ is $i$-the face of $f$.
Cohomology of this complex are called {\bf \blue singular
homology} of $M$.

\definition
{\bf \blue The group of $i$-cochains} is $C^i:=\Hom(C_i, \Z)$.
The boundary operator $\6:\; C_i\arrow C_{i-1}$
defines {\bf \blue the coboundary operator}
(denoted by the same letter) $\6:\; C^i\arrow C^{i-1}$.
The cohomology of the complex
\[ ... \stackrel \6 \arrow C^i \stackrel \6 \arrow C^{i+1}  \stackrel \6 \arrow...
\]
are called {\bf \blue singular cohomology of $M$}.

\definition
Let $R$ be a ring (typically, $R$ will be a field $\R$).
{\bf \blue The group of $i$-cochains with values in $R$} 
is $C^i_R:=\Hom_\Z(C_i, R)$, where $\Hom_\Z$ denotes
the homomorphism of abelian groups. Cohomology
of the corresponding cochain complex are called
{\bf \blue cohomology with coefficients in $R$}.

\newpage

{\bf \blue Singular cohomology}

In the next lecture, I will prove\\
\theorem {\bf \blue (de Rham theorem)}
Let $H^i_{dR}(M)$ denote de Rham cohomology of a manifold $M$
and $H^i(M, \R)$ denote the singular cohomology with
coefficients in $\R$. {\bf \blue Then $H^i_{dR}(M)=H^i(M,\R)$.}

\begin{center}
{\small \green 
To prove equivalence of different cohomology theories,\\ Eilenberg and MacLane
developed the theory of categories.}\\[1mm]


\epsfig{file=Eilenberg-MacLane.jpg,width=0.57\linewidth}\\[3mm]
{\small \green Saunders MacLane (1909-2005) and Samuel Eilenberg (1913-1998) }
\end{center}

\newpage

{\bf \blue Categories}

\newcommand{\Ob}{\operatorname{{\cal O}b}}
\newcommand{\Mor}{\operatorname{{\cal M}or}}

\definition 
{\bf \blue A category} $\cac$ is a collection of data
called ``objects'' and ``morphisms between objects''
which satisfies the axioms below.

{\bf \green DATA.}\\
\phantom{XX} {\bf \red Objects:} A class
$\Ob(\cac)$ of {\bf \blue objects}
of $\cac$. \\
\phantom{XX} {\bf \red Morphisms:} For each $X, Y\in \Ob(\cac)$,
one has a set $\Mor(X,Y)$ of {\bf \blue morphisms from $X$ to $Y$}.\\
\phantom{XX} {\bf \red Composition of morphisms:} For each
$\phi\in \Mor(X,Y), \psi \in \Mor(Y,Z)$ there exists
{\bf \blue the composition} $\phi\circ \psi \in \Mor(X, Z)$\\
\phantom{XX} {\bf \red Identity morphism:} For each 
$A\in \Ob(\cac)$ there exists a morphism $\Id_A \in \Mor(A,A)$.

{\bf \green AXIOMS.}\\
\phantom{XX} {\bf \red Associativity of composition:}
$\phi_1\circ(\phi_2\circ\phi_3)=(\phi_1\circ\phi_2)\circ\phi_3$.\\
\phantom{XX} {\bf \red Properties of identity morphism:}
For each $\phi\in \Mor(X,Y)$, one has
$\Id_x\circ \phi = \phi = \phi\circ \Id_Y$\\

\newpage

{\bf \blue Categories (2)}

\definition
Let $X, Y\in \Ob(\cac)$ be7objects of $\cac$.
A morphism $\phi\in \Mor(X,Y)$ is called {\bf \blue an isomorphism}
if there exists $\psi\in \Mor(Y,X)$ such that
$\phi \circ \psi = \Id_X$ and $\psi\circ\phi = \Id_Y$.
In this case, the objects $X$ and $Y$ are called
{\bf \blue isomorphic}.


{\bf \green Examples of categories:}\\
{\bf \purple Category of sets:} its morphisms are arbitrary maps.\\
{\bf \purple Category of vector spaces:} its morphisms are linear maps.\\
{\bf \purple Categories of rings, groups, fields:} morphisms are 
homomorphisms.\\
{\bf \purple Category of topological spaces:} morphisms are continuous maps.\\
{\bf \purple Category of smooth manifolds:} morphisms are smooth maps.

\newpage

{\bf \blue Functors}

\definition
Let $\cac_1, \cac_2$ be two categories.
A {\bf \blue covariant functor} from $\cac_1$ to
$\cac_2$ is the following set of data.

1. {\bf \purple A map $F:\; \Ob(\cac_1)\arrow \Ob(\cac_2)$}.

2. {\bf \purple A map $F:\; \Mor(X,Y) \arrow \Mor(F(X), F(Y))$
defined for any pair of objects $X, Y\in \Ob(\cac_1)$.}

These data define a functor if they are {\bf \purple compatible with
compositions,} that is, satisfy $F(\phi) \circ F(\psi) = F(\phi\circ\psi)$
for any $\phi\in \Mor(X,Y)$ and $\psi\in \Mor(Y,Z)$, and
{\bf \purple map identity morphism to identity} morphism.

\newpage

{\bf \blue Example of functors}

{\bf \purple A ``natural operation'' on mathematical objects is usually
a functor.} Examples:

1. A map $X \arrow 2^X$ from the set $X$ to the set of all subsets
of $X$ is a functor from the category $\Sets$ of sets to itself.

2. A map $M \arrow M^2$ mapping a topological space
to its product with itself is a functor on topological spaces.

3. A map $V \arrow V \oplus V$ is a functor on vector spaces;
same for a map $V \arrow V\otimes V$ or $V \arrow (V \oplus V)\otimes V$.

4. {\bf \blue Identity functor} from any category to itself.

5. A map from topological spaces to $\Sets$,
putting a topological space to the set of its connected components.

\exercise {\bf \purple Prove that it is a functor.}

\newpage

{\bf \blue Contravariant functors}

\definition
Let $\cac$ be a category. Define the 
{\bf \blue opposite category} $\cac^{op}$ with the same set of objects, and
 $\Mor_{\cac^{op}}(A,B)= \Mor_{\cac}(B,A)$. The composition
$\phi\circ \psi$ in $\cac$ gives
the composition $\psi^{op}\circ \phi^{op}$ in $\cac^{op}$.

\definition
A {\bf \blue contravariant functor} from $\cac_1$ to $\cac_2$
is the usual (``covariant'') functor from $\cac_1$ to $\cac_2^{op}$.

\example
A map from the category of topological spaces to category of rings
mapping a space to a ring of continuous functions on it gives
a contravariant functor.

\example
Let $X \in \Ob(\cac)$ be an object of $\cac$.
A map $Y\arrow \Mor(X,Y)$ defines a covariant functor
from $\cac$ to the category $\Sets$ of sets. A map
$Y \arrow \Mor(Y,X)$ defines a contravariant functor
from $\cac$ to $\Sets$. Such functors to $\Sets$
are called {\bf\blue representable}.

\newpage

{\bf \blue Equivalence of functors}

\definition
Let $X, Y\in \Ob(\cac)$ be objects of a category $\cac$.
A morphism  $\phi\in \Mor(X,Y)$ is called {\bf \blue an isomorphism}
if there exists $\psi\in \Mor(Y,X)$ such that
$\phi \circ \psi = \Id_X$ and $\psi\circ\phi = \Id_Y$.
In this case $X$ and $Y$ are called {\bf \blue isomorphic}.

\definition
Two functors  $F, G:\;\cac_1\arrow \cac_2$ 
are called {\bf\blue equivalent} if for any $X \in \Ob(\cac_1)$
we are given an isomorphism $\Psi_X:\; F(X) \arrow
G(X)$, in  such a way that for any  $\phi\in \Mor(X,Y)$,
one has $ F(\phi) \circ \Psi_Y= \Psi_X\circ G(\phi)$.

\remark
Such commutation relations are usually expressed by 
{\bf \blue commutative diagrams}. For example, the condition
$F(\phi) \circ \Psi_Y= \Psi_X\circ G(\phi)$ is expressed
by a commutative diagram
\begin{equation*}
\begin{CD}
F(X) @>{F(\phi)}>> F(Y)\\
@V{\Psi_X}VV @VV{\Psi_Y}V\\
G(X) @>{G(\phi)}>> G(Y)
\end{CD}
\end{equation*}
A collection of morphism  $\Psi_X:\; F(X) \arrow G(X)$
is called {\bf \blue a natural transform of functors}
if this diagram is commutative for all $\phi\in \Mor(X,Y)$.

\newpage

{\bf \blue Equivalence of cohomology: framework argument}

\remark All cohomology theories 
give a contravariant functor from 
topological spaces to abelian groups.

\theorem
Let $H_1^i, H_2^i$ be contravariant functors (``cohomology theories'')
from the category of topological spaces to vector spaces,
where $i=0, 1, 2, ... $.
Suppose that they satisfy the following axioms.

{\em \green
1. Both cohomology theories $H_1^*$ and $H_2^*$ 
have Mayer-Vietoris long exact sequences.

2. There is a natural transform of functors $H_1 \stackrel \Psi\arrow H_2$
inducing isomorphism on cohomology of a starlike set.}

{\bf \red Then $\Psi$ is an equivalence of functors.}

In this situation we also say that 
{\bf \blue cohomology theories $H_1^*$ and $H_2^*$ are equivalent.}

\end{document}
