First I'll go through my working. Throughout we assume the homology groups of the torus and circle are known.

Let $X=S^1 \times S^1$ be the torus, and $A=S^1 \times \{1\}$. The following is part of a long exact sequence: $$ \dots \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \to \dots $$ For $n\geq 3$, we have $H_n(X) = H_{n-1}(A) = 0$, so $H_n(X,A) = 0$. Using the 'reduced version' of the sequence, we have $$ \dots \to H_2(A) \to H_2(X) \to H_2(X,A) \xrightarrow{f} H_1(A) \xrightarrow{i_*} H_1(X)\qquad\qquad\qquad\qquad $$ $$ \qquad\qquad\qquad\qquad\qquad \to H_1(X,A)\to \tilde{H}_0(A) \to \tilde{H}_0(X) \to \tilde{H}_0(X,A) \to 0. $$

Step 1. Now, $A, X$ are path connected so $\tilde{H}_0(A) =\tilde{H}_0(X) = 0$. So $\tilde{H}_0(X,A)=0$ by the long exact sequence, which gives us that $H_0(X,A) \cong \mathbb{Z}$.

Step 2. The map $i_*: H_1(A) \to H_1(X)$ takes $[\alpha] \mapsto [i(\alpha)]$ where $i$ is the inclusion. In other words, it is injective, and so $0 = \ker i_* = \text{im}\:f$, and so $f$ is the zero map. This then leaves two short exact sequences, and implies that $H_2(X,A) \cong H_2(X) \cong \mathbb{Z}$.

Step 3. It is an immediate consequence of the other remaining short exact sequence that $H_1(X,A) \cong H_1(X)/\text{im }i_* \cong \mathbb{Z} \oplus \mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}$.

What I'm looking for are mainly comments on if this was done properly, but I also have two specific queries:

1. Have I used the reduced homology 'version' of the long exact sequence correctly? I'm very iffy with reduced homology...

2. My reason for $i_*$ being injective was a little handwavy. Is it always the case that a homomorphism induced by inclusion of spaces $A\hookrightarrow X$ is injective?


2 Answers 2


Overall, this looks pretty good. You probably want to point out that $H_2(A)\cong0$.


  1. Yes. You wrote the sequence down properly. Reduced homology is largely used to make computations easier. It arises from augmenting the (singular, simplicial, etc.) chain complex associated to a space with an additional map $\cdots\to C_0(X)\xrightarrow{\epsilon}\mathbb{Z}\to 0$ where $\epsilon(\sum_i n_i\sigma_i)=\sum_i n_i$.

  2. It is not always the case that the map induced by an inclusion of spaces is an injection on homology. Consider the inclusion $S^1\hookrightarrow D^2$. The disc is contractible, and so has the same homology as a point, where as $S^1$ has nontrivial $H_1$. In your case, you can see that the inclusion of $S^1$ into $S^1\times S^1$ as $S^1\times \{*\}$ induces an inclusion on homology by giving both spaces suitable cell-complex structures (be it CW-, $\Delta$-, or simplicial-) and explicitly following the map induced on the chain complex level to homology.

  • 2
    $\begingroup$ About 2: Since $i$ has a retraction $r:X\to A:(s,t)\mapsto (s,1)$, the map $i_{\ast}$ is injective by functoriality: $(r\circ i)_{\ast}=Id_{H_1(A)}=r_{\ast} \circ i_{\ast}$ $\endgroup$ Jun 5, 2011 at 18:50
  • $\begingroup$ @Fahad Sperinck Be sure to take @Georges Elencwajg's comment seriously! $\endgroup$
    – wckronholm
    Jun 6, 2011 at 15:58

I can't comment because I don't have enough points, but I'd like to remark that $\tilde H_n(X,A)$ is the same as $H_n(X,A)$ for all $n$ when $A\ne \emptyset$, so this step 1 conclusion seems incorrect...

  • 1
    $\begingroup$ Indeed $H_0(X,A)=0$ for any two path connected spaces $A,X$. $\endgroup$ Nov 3, 2013 at 23:30

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