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I have a space $Z=X\sqcup P$, where $P=\{*\}$ is a disjoint point. I want to show that $\tilde{H_i}(Z) \cong H_i(X)$. I start with the following: Let $f:A \rightarrow C$ and $g:B \rightarrow C$ be homomorphism of abelian groups where we also assume $g$ to be isomorphism. We then also define $f\oplus g:A\oplus B \rightarrow C$ as $(a,b) \mapsto f(a) + f(b)$. We can then construct an isomorphism $\omega$ between $A$ and the kernel of $f \oplus g$ by sending $a$ to $(a,g^{-1}(-f(a)))$. One can define the $n$-th reduced homology in the following way: let $X$ be a space and let $P = \{∗\}$ be a one-point space. Then there is a unique continuous map $\gamma^X:X \rightarrow P$. The $n$-th reduced homology group of $X$ is: \begin{equation} \tilde{H_i}(X)=ker(\gamma_{*}^X:H_n(X) \rightarrow H_n(P)). \end{equation} So I argue as follows: \begin{equation} \tilde{H_n}(Z)=ker(\gamma^Z:H_n(Z)\rightarrow H_n(P)) \cong ker(\gamma:H_n(X)\oplus H_n(P) \rightarrow H_n(P)). \end{equation} Let $f^{'}: H_n(X) \rightarrow H_{n}(P)$ be a group homomorphism (is that always well-define, can I do that?) and $g^{'}:H_n(P) \rightarrow H_n(P)$ an isomorphism (so $g^{'}$ is just $=$). Then we get $H_i(X) \cong ker(f^{'} \oplus g^{'}) \cong ker(\gamma)\cong \tilde{H_n}(Z)$. Does this make sense at all?

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    $\begingroup$ It is not clear to me what you are doing with these homomorphisms. Why are you not defining them? Perhaps you should use the fact there is a right inverse to the map $X \sqcup *\rightarrow *$ given by including the point as the disjoint point. Analyze what happens on homology. $\endgroup$ Commented Dec 20, 2019 at 2:23
  • $\begingroup$ Thank @ConnorMalin. I am not sure I understand, I am still really new to this stuff and struggle to work things out from definitions. So, are you saying that $X \sqcup *$ are homeomorphic and so this induces the isomorphism between the homology groups of $X \sqcup *$ and $*$? And I should somehow go from there? $\endgroup$
    – billy192
    Commented Dec 20, 2019 at 11:11

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