# Hatcher's proof of Quillen plus construction

This is from Hatcher's Algebraic Topology. This Proposition says we can kill the fundamental group of an arbitrary connected CW complex with $$H_1=0$$, without affecting the homology groups. The proof was clear to understand, until the underlined statement. I see that $$H_2(X')\cong \pi_2(X')$$ by the Hurewicz theorem , but how can we represent a basis of $$H_2(X',X)$$ by maps $$\psi_\alpha :S^2 \to X'$$?

You have a typo in your very last formula, it should be $$\psi_\alpha : S^2 \to X'$$ (not $$\psi_\alpha : S^2 \to X$$).
The Hurewicz homomorphism $$\mathcal H : \pi_2(X') \approx H_2(X')$$ (which is an isomorphism in this case where $$X'$$ is simply connected) is defined in the statement of the Hurewicz theorem. The definition is this: given an element $$[\psi] \in \pi_2(X')$$ represented by a continuous map $$\psi : S^2 \to X'$$, the homology class $$\mathcal H[\psi] \in H_2(X')$$ is defined to be the image of the standard generator of $$H_2(S^2)$$ under the induced homomorphism $$\psi_* : H_2(S^2) \to H_2(X')$$. Since $$\mathcal H$$ is an isomorphism in this case, it follows that every element of $$H_2(X')$$ can be represented in the form $$\mathcal H[\psi]$$ for some appropriate choice of $$\psi$$.
So, pick $$\{\alpha_i\}$$ to be a basis for $$H_2(X',X)$$, let $$\{\alpha'_i\} \subset H_2(X')$$ be the image of that basis under the splitting, and so you may choose $$\psi_{\alpha_i} : S^2 \to X'$$ which represents $$\alpha_i$$ under the Hurewicz homomorphism as just described.