Take long exact sequence (l.e.s.) $H_1(A)\to H_1(X)\to H_1(X,A)\to H_0(A)\to H_0(X)\to H_0(X,A)\to 0$.

Suppose $H_1(X,A)=0$. Then $f:H_1(A)\to H_1(X)$ is surjective and $g:H_0(A)\to H_0(X)$ is injective. This is what I can deduce from the l.e.s. But I don't understand how to relate this phenomenon with path components of $X$. I don't understand how surjectivity of $f$ is related to path components of $X$. I looked at the following post

$H_1(X,A) = 0 \iff H_1(A) \rightarrow H_1(X)$ surjective and $X_i$ contains no more than one path-component of $A$

but I don't understand it; for example the part: $X_i$ path connected $\implies H_0(X_i) \cong \mathbb{Z}$. If $X_i$ contained more than one path-component $A_i$, say $n$, then $\operatorname{im}{ g} \cong \mathbb{Z}^n$, which is a contradiction to $\operatorname{im}{ g} \subset H_0(X_i) = \mathbb{Z}$.

I also don't see how a beginner who has just studied the portions from Hatcher's could come up with that.

So how to deduce using $f$ and $g$ that $X$ can't contain more that $1$ path component of $A$?

My other question here $H_0(X,A)=0$ iff $A$ intersects every path component of $X$. got closed. I think the following could work as a solution of that but I am not sure. Here, I used only definition of $H_0(X,A)$ and some intuition:

$H_0(X,A)=C_0(X,A)/\{\sigma(1)-\sigma(0): \sigma:[0,1]\to X, Im \sigma\not\subset A\}$

$=FAB\{\sigma: \Delta^0\to X, Im \sigma \not\subset A\}/\{\sigma(1)-\sigma(0): \sigma:[0,1]\to X, Im \sigma\not\subset A\}$

$=FAB\{\text{ path components of $X$ not intersecting $A$}\}$, here FAB means free Abelian group generated by.

The above working seems intuitive but I can't justify the last equality yet.

Thanks for your time.


1 Answer 1


First, we need to establish that $H_0(X)$ free abelian with basis in one-to-one correspondence with the path components of $X$. This is a straightforward exercise from the definition of $H_0$: The zero cycles are a free abelian group with basis given by the points $x \in X$, and the boundaries are freely generated by the (formal) differences $x - y$ for any $x, y$ that are in the same path component of $X$.

Now, the map $H_0(A) \to H_0(X)$ in the l.e.s. is not just any old group homomorphism. It comes from the inclusion map $A \hookrightarrow X$. For any basis element $[a] \in H_0(A)$, coming from a point $a \in A$, the image of $[a]$ in $H_0(X)$ is again $[a]$, coming from the point $a \in X$. Thus the condition that $H_0(A) \to H_0(X)$ is injective is equivalent to asserting that no two path components of $A$ are contained in a single path component of $X$.

Explicitly, suppose that $a_1, a_2 \in A$ are in different path components of $A$, but the same path component of $X$. Then $[a_1], [a_2] \in H_0(A)$ are distinct homology classes, but $[a_1], [a_2] \in H_0(X)$ are the same homology class. So $H_0(A) \to H_0(X)$ is not injective. On the other hand, if every path component of $A$ is contained in a different path component of $X$, then the basis elements $[a] \in H_0(A)$ all map to distinct basis elements of $H_0(X)$. Thus the map $H_0(A) \to H_0(X)$ is injective.

  • $\begingroup$ Thanks a lot for the answer. I understood until the last sentence. Can you please explain why injectivity of the map is equivalent to requiring that no two path components of A are contained in X? $\endgroup$
    – Koro
    Mar 28, 2023 at 5:15
  • $\begingroup$ Added an explicit argument for the last part. $\endgroup$
    – Alex G.
    Mar 28, 2023 at 14:13
  • $\begingroup$ Thanks a lot :-). $\endgroup$
    – Koro
    Mar 28, 2023 at 18:06

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