Homology and cofibrations

Question

Please help me to answer my question.It is finding $H_*(T^2,\lbrace\ast\rbrace\times S^1\cup S^1\times\lbrace\ast\rbrace)$ using cofibrations matter.I would be grateful for your answer.

Answer 1

I'll expand on user8268's answer a little.

With the usual cellular decomposition of the torus $T^2$, it is immediate (by the fact that cellular inclusions are cofibrations) that the inclusion $i\colon X\rightarrow T^2$ is a cofibration.

Now, if the inclusion of a subspace is a cofibration, then the relative homology of the pair is isomorphic to the homology of the quotient and so we have $$H_*(T^2,X)\cong H_*(T^2/X).$$

Now, to see that $T^2/X$ is homeomorphic to a $2$-sphere, it's easiest to visualise by constructing the torus as the quotient of a unit square with the usual side pairings via the quotient map $q\colon I^2=I\times I\rightarrow T^2$. If $i'\colon\partial I^2\rightarrow I^2$ is the inclusion of the boundary of the square, then we see that $$q\circ i'=i\circ q|_{\partial I^2}$$ where $q|_{\partial I^2}\colon\partial I^2\rightarrow X$ is the quotient map $q$ restricted to the boundary of the square.

Let $\tilde{q}\colon T^2\rightarrow T^2/X$ be the canonical quotient map, then $\tilde{q}\circ q$ is also a quotient map and by the above commutativity of maps is a quotient of $I$ by the subspace $\partial I^2$. It is clear that $I^2/\partial I^2\cong S^2$ and so $T^2/ X\cong S^2$.

Finally it follows that $H_*(T^2,X)\cong H_*(S^2)$ which is free abelian on one generator in degrees $0$ and $2$, and trivial in all other degrees.

Answer 2

Since $X\subset T^2$ ($X=\lbrace\ast\rbrace\times S^1\cup S^1\times\lbrace\ast\rbrace$) is a cofibration, $H_*(T,X)$ is the reduced homology of $T/X$. As $T/X\cong S^2$, it is $\mathbb Z$ in dimension 2, and 0 elsewhere.