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

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.

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