Hatcher Exercise 2.1.16
(a) Theorem: if and only if meets each path component of .
Proof: We start with a lemma, and then apply Proposition 2.6 from Hatcher.
Lemma: For with path-connected, is trivial.
Proof: From the long exact sequence relating homology groups and relative homology groups, we have:
is generated by a single basepoint – in particular, by some basepoint . Thus, is generated by , and is at least among the generators of ( may not be path connected itself). Then, we must at least have that is surjective. But this means is , and .
Now, we use the fact that for a general space , there is an isomorphism for each path component of . Let be the non-empty intersection of with each path component of . Then, we have:
The boundary maps in this sequence all split across the direct sums, and we apply the lemma (note each is path-connected) to each component, yielding for all . Then, as well.
(b) Theorem: if and only if is surjective and each path-component of contains at most one path component of .
Proof: Again, start with a lemma:
Lemma: For with path-connected, is trivial if and only if is surjective and is path-connected.
Proof: Consider the relevant portion of the long exact sequence:
From a previous result, we know that is trivial if and only if is surjective (a hypothesis), and is injective. We show the latter.
Note that is generated by a single basepoint, in particular a basepoint in . Then, is generated by the same point (here we use that is path-connected). Thus, in fact, and is clearly injective.
Consider now a general space with path components , and let . Then, again we consider the exact sequence:
The boundary maps split across the direct sums, and we apply the lemma to each component (this time, note that each is path-connected because each path component of contains no more than one path component of ). Then, we conclude is trivial for all , and thus is trivial.