# Show that $H_1(X,A) = 0$ iff $H_1(A)→H_1(X)$ is surjective and each path-component of X contains at most one path-component of A.

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.

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.

• 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?
– Koro
Mar 28, 2023 at 5:15
• Added an explicit argument for the last part. Mar 28, 2023 at 14:13
• Thanks a lot :-).
– Koro
Mar 28, 2023 at 18:06