# path connected subspace $A$ of $X$, $i:A\to X$ inclusion. Why is the induced map of $i$ on homology injective?

X is a topological space and $A\subset X$ is a path-connected subspace of X and $i:A\to X$ is the inclusion. I want to know, why the induced map of i on singular homology of dimension zero over $\mathbb{Z}$, namely $H_0(i)=(i)_0:H_0(A)\to H_0(X)\; [v]\mapsto [i(v)]$, is injective.
My first thoughts are: if i consider the pair $(X,A)$, i have a long exact sequence
$$\ldots\to H_{n+1}(X,A)\overset{\delta_n+1}{\to}H_n(A)\overset{(i)_n}{\to}H_n(X)\overset{(j)_n}{\to}H_n(X,A)\overset{\delta_n}{\to}H_{n-1}(A)\to\ldots$$
with $\delta_n$is the boundary map and $j:(X,\emptyset)\to (X,A)$ is the inclusion map on pairs and $(j)_*$ is the induced map of j. I consider the section of the sequence:
$$\ldots H_1(X)\overset{(j)_1}{\to} H_{1}(X,A)\overset{\delta_1}{\to}H_0(A)\overset{(i)_0}{\to}H_0(X)\overset{(j)_0}{\to}H_0(X,A)\overset{\delta_0}{\to}0$$.
Why $(i)_0$ is injective? I only know: $(j)_0$ is surjective, because $Im(j)_0=H_0(X,A)=ker(\delta)_0$ But i need to know, why $Im(\delta_1)=ker(i)_0$ has to be zero, but i don't know why. Could you help me?
Furthermore i know: $H_0(A)\cong\mathbb{Z}$ and $H_0(X)\cong \mathbb{Z}^{\text{|path-connected-components of X|}}$. But i don't know how to continue...Regards

Sometimes, you actually have to look at the topology.

A chain $c$ in $H_1(X, A)$ is a collection of edges (with coefficients, but those will turn out to be irrelevant); the boundary is a collection of pairs-of-points in $A$. Since $A$ is path connected, for each such pair we can find a path in $A$ that connects them. The sum of all these paths, with the appropriate coefficients, has boundary equal to $\partial c$. Hence $\partial c = 0$ as an element of the homology group $H_0(X, A)$. Since the map before it in the exact sequence is a zero map, $(i)_0$ must be injective.

Small hint: nowhere in your reasoning did you use the path-connectedness of $A$, so it was clear to me that it had to be part of the story. (Esp. when you consider the case where $X$ is the unit interval and $A = \{0, 1\}$ is its boundary: in this case, the map is not injective, but that's because $A$ is not connected.)

• thank you. i have a question. what do you mean with "pairs-of-points"? i have problems with the boundary map $H_1(X,A)\to H_0(A)$ from the snake lemma (to understand this map). But the main argumentation here seems to be reasonable for me. – Sabrina G. Nov 25 '14 at 17:32
• Take a single 1-simplex $c : I \to X$; its boundary is the 0-chain $c(1) - c(0)$. We can think of that as a pair of points in $X$ (one with a coeff of $+1$, one with a coeff of $-1$). For a more general chain, you get multiple simplices, with different coefficients, but each contributes a "pair of points" like this. – John Hughes Nov 25 '14 at 19:27
• ah okay. i was not sure if you mean something different (something in relation to a pair of spaces) but now i understand. Thank you – Sabrina G. Nov 25 '14 at 19:34

The long exact sequence is not really helpful here. What we need to do is look at chain groups. Any two 0-chains in $A$ of the form $1f$ where $f$ is a constant map are homologous and map to a generator when we pass to homology. 0 chains of the form $1(i\cdot f)$ generate the direct summand in $H_0(X)$ corresponding to the path component containing $A$. Therefore a generator of $H_0(A)$ is mapped to a nonzero element in $H_0(X)$, so since the groups are free abelian the map is injective.

• hi. yesterday i was tired.. could you explain the last sentence/ the last conclusion, why the map is injective if the generator of $H_0(A)$ is mapped to a nonzero element in $H_0(X)$ and the groups are free abelian? – Sabrina G. Nov 26 '14 at 9:07
• @Teddy Specifically the generator is mapped to a generator of a direct summand, but any nonzero element generates an infinite cyclic subgroup. If the map were not injective the image would be finite. – Matt Samuel Nov 26 '14 at 14:07