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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?

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  • $\begingroup$ 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
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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.

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