My question is about what one could say about the Betti number of both spaces $X$ and $Y$ relative to one another if we have a map $f$ between them (e.g., a classical case is when $f$ is a covering map). Is there an inequality if $f$ happens to be injective or surjective?

Thank you in advance.

  • $\begingroup$ related: math.stackexchange.com/questions/41451/… $\endgroup$ – Grigory M May 29 '11 at 8:52
  • $\begingroup$ Thanks Grigory, the answer was interesting it says that if the map is injective with a section then it induces an injective map of homology modules which means that the betti numbers of X are less or equal to those of Y. I am wondering if there is an answer to my question in a more general setting. $\endgroup$ – El Moro May 29 '11 at 8:57
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    $\begingroup$ The morale from the linked question: injectivity/surjectivity of $f$ implies no inequality for Betty numbers whatsoever. So it's not quite clear what kind of answer you expect. $\endgroup$ – Grigory M May 29 '11 at 9:02
  • $\begingroup$ I am actually looking for those special cases (like the one you mentioned with a section) in which one could compare betti numbers. $\endgroup$ – El Moro May 29 '11 at 9:03

There is no relation whatsoever, basically.

  • Every manifold embeds in a contractible space ($\mathbb{R}^N$ for big enough $N$), and every nonempty manifold contains a contractible subspace. Of course manifolds can have all sorts of Betti numbers, while all the Betti numbers of a contractible are zero (except for $b_0 = 1$).

  • Any nonempty space always surjects onto a singleton. Meanwhile, for any path-connected space $X$ with a chosen $x \in X$, $PX = \{ \gamma : [0,1] \to X : \gamma(0) = x \}$ is contractible (collapse the interval to a point), and it surjects onto $X$ via $\gamma \mapsto \gamma(1)$.

So that pretty much rules out any connection between the existence of a map $f$ (even injective or surjective) and the Betti numbers. Even a bijective continuous map is not required to preserve Betti numbers: think about $[0,1) \to S^1$, $t \mapsto e^{2i\pi t}$, for example.

The only thing I can think of is that if there exists a maps $f : X \to Y$ and $b_0(X) \ge 1$ then $b_0(Y) \ge 1$ (a not very deep statement about maps between empty and nonempty spaces). Similarly you can craft something relating surjections and connected components (if $f : X \to Y$ is a surjection then $b_0(Y) \le b_0(X)$) but that's it, and that's not very insightful.

If you know that the surjection (resp. injection) is split, though, you can say something. Indeed suppose that you have $f : X \to Y$, $g : Y \to X$ and $g \circ f = \operatorname{id}_X$. Then $g_* \circ f_* = \operatorname{id}_{H_* X}$, and so you do get an inequality on Betta numbers $b_i(X) \le b_i(Y)$.

Homotopically, "injection" and "surjection" are not well-behaved notions. In general you want to deal with their homotopy "analogues", cofibrations and fibrations.

Something that is true is that if you have a fibration $p : E \to B$ with homotopy fiber $F$, then $\chi(E) = \chi(B) \chi(F)$ where $\chi = \sum \pm b_i$ is the Euler characteristic. This doesn't give you an inequality straight away on Betti numbers, but at least you can say something.

When you have a cofibration, you can also say something. If $A \to X$ is a cofibration, you can view $A$ as a subspace of $X$, and then it's a general fact about cofibrations that $H_n(X/A) \cong H_n(X, A)$. Thus $\chi(A) + \chi(X/A) = \chi(X)$ by the long exact sequence in homology of a pair. However the Euler characteristic is not always nonnegative, of course, so you can't get a straightforward inequality like $\chi(X) \ge \chi(A)$, but you can still say things.

It's possible to apply both examples to a general map $f : X \to Y$, by replacing it either by a fibration or a cofibration. Then if $F$ is the homotopy fiber of $f$ and $C$ is the homotopy cofiber of $f$, then you still have $\chi(X) = \chi(Y) \chi(F)$ and $\chi(X) + \chi(C) = \chi(Y)$. But you really have to take replacements here, the usual fiber or cofiber wouldn't work (look at any example that is either not a fibration or a cofibration).

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    $\begingroup$ Not sure if that's what the OP wanted (and OP hasn't been seen since June 2011 anywya), but this question has been unanswered for 3.5 years and has just been bumped onto the front page, so here goes. $\endgroup$ – Najib Idrissi Jan 10 '15 at 12:15

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