comparing Betti numbers 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.
 A: There is no relation whatsoever, basically.


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*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).
