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.

Thanks for reading.

  • $\begingroup$ Use the long exact sequence for reduced homology groups. $\endgroup$
    – Paul Frost
    May 29 '21 at 22:32
  • $\begingroup$ @Paul Frost get it now. Thanks. $\endgroup$
    – Anacardium
    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 $H^0$.

  • $\begingroup$ The generators of $H_0(X)$ are the path components of $X.$ $\endgroup$
    – Anacardium
    May 29 '21 at 17:16
  • $\begingroup$ @Anacardium Do you understand why? $\endgroup$ May 29 '21 at 17:18
  • $\begingroup$ 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. $\endgroup$
    – Anacardium
    May 29 '21 at 17:24
  • $\begingroup$ 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 $X.$ $\endgroup$
    – Anacardium
    May 29 '21 at 17:31
  • $\begingroup$ @Anacardium Right, and with that in mind you can just use the definition of your map to see where the two generators go. $\endgroup$ May 29 '21 at 17:32

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