# Computing relative homology group of the standard $1$-simplex with respect to it's boundary.

Find the relative homology group $$H_* ([0,1], \{0,1\}).$$

Since $$[0,1]$$ is contractible it follows from homotopy invariance property that $$H_n ([0,1]) = \begin{cases} \Bbb Z & n = 0 \\ 0 & n \geq 1 \end{cases}$$ and since $$\{0,1\}$$ is discrete it follows that $$H_*(\{0,1\}) = H_*(\{0\}) \oplus H_*(\{1\}).$$ So we have $$H_n (\{0,1\}) = \begin{cases} \Bbb Z \oplus \Bbb Z & n = 0 \\ 0 & n \geq 1 \end{cases}$$ Now we have the following short exact sequence of chain complexes $$0 \rightarrow C_*(\{0,1\}) \rightarrow C_*([0,1]) \rightarrow C_*([0,1], \{0,1\}) \rightarrow 0$$ This will induce a long exact sequence of the homology groups $$\cdots \rightarrow H_n(\{0,1\}) \rightarrow H_n([0,1]) \rightarrow H_n([0,1], \{0,1\}) \rightarrow H_{n-1}(\{0,1\}) \rightarrow H_{n-1} [0,1] \rightarrow \cdots$$

Using exactness of the above long exact sequence and from the fact that higher homology groups (except the $$0$$-th one) of $$[0,1]$$ are all trivial it is easy to show that for $$n \geq 2$$ $$H_n ([0,1], \{0,1\}) = 0.$$ But I find it difficult to find $$0$$-th and the $$1$$-th relative homology groups. Since $$H_1(\{0,1\}) = 0$$ we find that $$H_1 ([0,1])$$ is a subgroup of $$H_1([0,1],\{0,1\}).$$ In fact it is a proper subgroup as there is no injective homomorphism from from $$\Bbb Z \oplus \Bbb Z$$ ($$= H_0(\{0,1\})$$) to $$\Bbb Z$$ ($$= H_0([0,1])$$). But I can't conclude anything more than that. A small hint will be warmly appreciated at this stage.

• Use the long exact sequence for reduced homology groups. May 29 '21 at 22:32
• @Paul Frost get it now. Thanks. May 30 '21 at 4:55

Hint: You can find how your map $$\Bbb Z \oplus \Bbb Z \to \Bbb Z$$ looks explicitly, if you know why the groups are of this form. What are the generators of $$H_0(X)$$?
$$H^0([0,1])$$ is generated by points of $$[0,1]$$ regarded as $$0$$-simplices, all of which are, as you pointed out, homological to each other (i.e. differ by an element of $$im \partial$$). Both summands in $$\Bbb Z \oplus \Bbb Z$$ are represented by points and so are mapped to a generator of $$H^0([0,1]),$$ so your map has (depending on orientation you choose) the matrix $$( \pm1, \pm1)$$. It's easy to see that its kernel is $$\Bbb Z,$$ so this is (by exactness) your $$H^0$$.
• The generators of $H_0(X)$ are the path components of $X.$ May 29 '21 at 17:16
• Yes of course I understand. Since $H_0(X) = C_0(X)/ \text {Im}\ \partial.$ Now $C_0(X) = \Bbb Z \{\textbf {Maps} (\Delta^0 , X)\} = \Bbb Z \{x \in X\}$ and $\text {Im}\ \partial$ is the free abelian group generated by the paths in $X.$ So the result follows. May 29 '21 at 17:24
• Sorry I mean $\text {Im}\ \partial$ is the free abelian group generated by the difference between initial and final point of the paths in $X.$ So if we quotient it out then all the points in the same path components are getting identified and hence the resultant group will be a free abelian group generated by the path components of $X.$ May 29 '21 at 17:31