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Homology and cohomology operators
Redirect to:
Homology
Given a cell map $f \colon K \rightarrow L$, the homology operators (or map) induced by f, $$f_* \colon H_k(K) \rightarrow H_k(L), {\rm \hspace{3pt} for \hspace{3pt} all \hspace{3pt}} k,$$ is a linear operator given by $$f_*([x]) = [f_k(x)],$$ where $$f_k \colon C_k(K) \rightarrow C_k(L)$$ is the chain map induced by $f$, and $[.]$ stands for the homology class.
The following is obvious.
Theorem 1. The identity map induces the identity homology operator: $$({\rm id}_{|K|})_* = {\rm id}_{H(K)}.$$
This is what we know about compositions of cell maps:
Theorem 2. The homology operator of the composition is the composition of the homology operators $$(gf)_* = g_*f_*.$$
The main result related to this theorem is below.
Theorem 3. Suppose K and L are cell complexes, if a map $$f \colon |K| \rightarrow |L|$$ is a cell map and a homeomorphism, and $$f^{-1} \colon |L| \rightarrow |K|$$ is a cell map too, then the homology operator $$f_* \colon H_k(K) \rightarrow H_k(L)$$ is an isomorphism for all $k$.
Proof. From the definition of inverse function, $$ff^{-1} = {\rm id}_{|L|},$$ $$f^{-1}f = {\rm id}_{|K|}.$$ From the above theorems, $$f_*(f^{-1})_* = {\rm id}_{H(L)},$$ $$(f^{-1})_* f_* = {\rm id}_{H(K)}.$$ $\blacksquare$
To summarize: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & & K & \ra{f} & L & & & H_*(K) & \ra{f_*} & H_*(L) & &\\ & cell \hspace{3pt} maps: & & \searrow ^{gf} & \da{g} & \leadsto & homology \hspace{3pt} operators: & & \searrow ^{g_*f_*} & \da{g_*}\\ & & & & M & & & & & H_*(M) & \end{array} $$
This is the realm of category theory.
Corollary. Under the conditions of the theorem, $$(f^{-1})_* = (f_*)^{-1}.$$
Theorem. If two maps are homotopic, they induce the same homology operators: $$f \sim g \Rightarrow f_* = g_*.$$
Cohomology
Cohomology theory seems very similar. In fact, the cohomology is isomorphic to homology as vector spaces, because they are dual.
It would seem natural to assume then that the map will induce the same linear operators. Just raise the indices, like this:
Given a cell map $f \colon K \rightarrow L$, the cohomology operator induced by $f$, $$f^* \colon H^k(K) \rightarrow H^k(L), {\rm \hspace{3pt} for \hspace{3pt} all \hspace{3pt}} k,$$ is a linear operator given by $$f^*([x]) = [f^k(x)],$$ where $$f^k \colon C^k(K) \rightarrow C^k(L)$$ is the cochain operator induced by $f$, and $[.]$ stands for the cohomology class.
Not so simple! We know that the last diagram is reversed...
Given a cell map $f \colon K \rightarrow L$, the cohomology operators (or map) induced by $f$, $$f_* \colon H_k(K) \rightarrow H_k(L), {\rm \hspace{3pt} for \hspace{3pt} all \hspace{3pt}} k,$$ is a linear operator given by $$f^*([x]) = [f^k(x)],$$ where $$f^k \colon C^k(L) \rightarrow C^k(K)$$ is the cochain map induced by $f$, and $[.]$ stands for the cohomology class.
The following is obvious.
Theorem 1. The identity map induces the identity cohomology operator: $$({\rm id}_{|K|})^* = {\rm id}_{H^*(K)}.$$
This is what we know about compositions of cell maps:
Theorem 2. The homology operator of the composition is the composition of the homology operators $$(gf)^* = f^*g^*.$$
The main result related to this theorem is below.
Theorem 3. Suppose K and L are cell complexes, if a map $$f \colon |K| \rightarrow |L|$$ is a cell map and a homeomorphism, and $$f^{-1} \colon |L| \rightarrow |K|$$ is a cell map too, then the cohomology operator $$f^* \colon H^k(L) \rightarrow H^k(K)$$ is an isomorphism for all $k$.
Proof. From the definition of inverse function, $$ff^{-1} = {\rm id}_{|L|},$$ $$f^{-1}f = {\rm id}_{|K|}.$$ From the above theorems, $$(f^{-1})^* f^* = {\rm id}_{H^*(L)}.$$ $$f^*(f^{-1})^* = {\rm id}_{H^*(K)},$$ $\blacksquare$
To summarize: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & & K & \ra{f} & L & & & H^*(K) & \la{f_*} & H^*(L) & &\\ & cell \hspace{3pt} maps: & & \searrow ^{gf} & \da{g} & \leadsto & cohomology \hspace{3pt} operators: & & \nwarrow ^{f^*g^*} & \ua{g^*}\\ & & & & M & & & & & H^*(M) & \end{array} $$
Corollary. Under the conditions of the theorem, $$(f^{-1})^* = (f^*)^{-1}.$$
Theorem. If two maps are homotopic, they induce the same cohomology operators: $$f \sim g \Rightarrow f^* = g^*.$$
