Given a path connected space $X$ and $x_0, x_1 \in X$, knowing that $H_0(X) \cong \mathbb{Z}$, show that $\operatorname{cls} \ x_0 = \operatorname{cls} \ x_1$ is a generator of $H_0(X)$

This was my attempt at a proof:

Proof: The map $\psi : H_0(X) \to \mathbb{Z}$ defined by $\psi\left(\sum m_i x_i + B_0(X)\right) = \sum m_i$ is an isomorphism.

Now choose $x_0, x_1 \in X$ and note that $\operatorname{cls} x_0 = x_0 +B_0(X)$ and $\operatorname{cls} x_1 = x_1 + B_0(X)$. Now $$\psi(x_0 + B_0(X)) = 1 = \psi(x_1 + B_0(X))$$ so that $\operatorname{cls} x_0 = \operatorname{cls} x_1$ by injectivity of the isomorphism. Furthermore since $H_0(X) \cong \mathbb{Z}$ and any isomorphism between cyclic groups takes generators to generators we must have that $\operatorname{cls} x_0 = \operatorname{cls} x_1$ are generators of $H_0(X)$. $\square$

Is my proof correct? The reason I ask this is because a slightly longer and different proof was given in class of this fact.

  • $\begingroup$ I agree with the proof. Is there any particular part you doubt? $\endgroup$ – Benny Zack Nov 10 '18 at 12:12

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