Let $X, Y$ be topological spaces. What are the possible maps $H_0(X) → H_0(Y)$ on homology coming from continuous maps $X → Y$? For example, can a map $X → X$ on a connected space induce a non-identity automorphism $H_0(X) → H_0(X)$?

My thoughts: Let $f\colon X → Y$ be continuous. For each path component $A$ of $X$ and $B$ of $Y$, select a base point $\star_A ∈ A$ and $\star_B ∈ B$. Then $$\star_A → A \overset {f\lvert_A} → Y → \star_B$$ induces an isomorphism $H_0(\star_A) → H_0(\star_B)$ and I guess that means that $H_0(f\lvert_A)$ is either an isomorphism or the null map, depending on whether $f$ maps $A$ into $B$ or not. Is this correct?

What (else) can I conclude about the nature of $H_0(f)$? If $X$ and $Y$ both have only finitely many components, does it look like a matrix whose only entries are $0$ and $1$? If $X = Y$ is connected, are non-identity automorphisms in homology possible?

  • $\begingroup$ The constant map $Y\to *_Y$ is not aware of any connected component to which $*_Y$ "really" belongs $\endgroup$ – Hagen von Eitzen Jan 31 '15 at 21:22
  • $\begingroup$ @HagenvonEitzen That’s true, so my intuition is flawed/plain wrong. $\endgroup$ – k.stm Jan 31 '15 at 21:24
  • $\begingroup$ @HagenvonEitzen I‘ve updated my reasoning and I hope that I have now correctly captured what I was vaguely thinking of. Does it make sense? $\endgroup$ – k.stm Jan 31 '15 at 21:35
  • $\begingroup$ In what you wrote you wrote component where you probably meant path component. $\endgroup$ – Mariano Suárez-Álvarez Jan 31 '15 at 22:14

By the definition of singular homology, $H_0 (X)$ is the abelian group generated by the points of $X$, modulo the equation $x_0 = x_1$ for every path connecting $x_0$ to $x_1$. As such, $H_0 (X)$ is the free abelian group generated by the path components of $X$. Given a continuous map $f : X \to Y$, it is not hard to see that the induced homomorphism $f_* : H_0 (X) \to H_0 (Y)$ must send generators to generators. Thus, the matrix corresponding to $f_*$ is indeed a zero-one matrix – in fact, it has the property that every column contains exactly one $1$.

Now suppose $X$ is locally path connected. Then by considering continuous maps which are constant on path components, it is not hard to see that every homomorphism $H_0 (X) \to H_0 (Y)$ corresponding to a zero-one matrix with exactly one $1$ in each column can be realised by a continuous map $X \to Y$.

In particular, if $X$ is path connected, then the only automorphism of $H_0 (X)$ coming from a continuous map is the identity; in other words, $- \mathrm{id} : H_0 (X) \to H_0 (X)$ does not come from any continuous map $X \to X$.

The real point is that the functor $H_0 : \mathbf{Top} \to \mathbf{Ab}$ factors as $\pi_0 : \mathbf{Top} \to \mathbf{Set}$ (the functor sending each topological space to its set of path components) followed by the free abelian group functor $\mathbf{Set} \to \mathbf{Ab}$. So your question is just asking which homomorphisms of free abelian groups are induced by a map between bases.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.