Let $X$ and $Y$ be CW-complexes and $\varphi\colon X\to Y$ be a cellular map. How can we describe the induced morphisms of cellular homology and cohomology groups explicitly? I suppose that $\varphi_*$ maps the homology class of a $n$-dimensional cell $e^n \subset X$ to the sum of the homology classes of $n$-dimensional cells $c^n_i \subset Y$ taken with integer coefficients which equal to the degrees of the maps $\psi_i \colon e^n/\partial e^n \to c^n_i/\partial c^n_i$. Is this description right?
Asked
Active
Viewed 157 times
2 Answers
1
Absolutely!
And it can be seen from Hatcher's definition of cellular homologies. In his notation cells are generators of $H_n(X_n, X_{n-1})$. So the map of pairs $f_{red}: (X_n/X_{n-1}, pt) \rightarrow (Y_n/Y_{n-1}, pt)$ is just a map between wedges of spheres.
FlyingHom
- 215
1
Your description is almost but not quite right.
To see why, consider the composition $$\partial e^n \xrightarrow{\chi_e \, | \, \partial e^n} X^{n-1} \xrightarrow{\phi} Y^{n-1} $$ where $\chi_e$ is the characteristic map of $e^n$ (I'm cheating a little bit, by thinking of the domain of this map as being homeomorphic to the closed $n$-disc). The problem is that $\phi(\chi_e(\partial e^n))$ is not a subset of $\partial c^n_i$, so it is not set-theoretically clear how to obtain an induced map $e^n / \partial e^n \to c^n_i / \partial c^n_i$.
What's better, instead of using $c^n_i / \partial c^n_i$, is to use $Y^n / (Y^n - \partial c^n_i)$, and you obtain this map which is induced from $\phi \circ \chi_e$: $$\psi_i : e^n / \partial e^n \to Y^n / (Y^n - \text{interior}(c^n_i)) $$ It pretty much amounts to the same thing as what you wrote, the image of this map in both cases is homeomorphic to $S^n$, but this formula actually makes sense and yours does not quite make sense.
Lee Mosher
- 120,280