Morphism between free abelian groups is 1-1; Vick Prop. 1.9

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Proposition 1.6 in Vick's Homology Theory states:

If $X$ is a space and $\{X_{\alpha}:\alpha \in A\}$ are the path components of $X$, then

$$H_{k}(X) \approx \sum_{\alpha \in A}H_{k}(X_{\alpha}).$$

The first step of the proof states that there is a "natural" homomorphism

$$\Psi: \sum_{\alpha \in A}S_{k}(X_{\alpha}) \longrightarrow S_{k}(X)$$

given by

$$\Psi((\sum_{\phi_{\alpha}}n_{\phi,\alpha}\cdot \phi_{\alpha}: \alpha \in A) = \sum_{\alpha \in A}(\sum_{\phi_{\alpha}}n_{\phi,\alpha}\cdot \phi_{\alpha}),$$

and that $\Psi$ must be a monomorphism since the groups involved are free abelian. As far as I understand, a morphism between free abelian groups need not be 1-1---am I missing something in the discussion so far that makes this assertment of injectivity true? Other than working it out directly, of course. I have a very light understanding of natural morphisms and I'm not sure if "natural" here means algebraically natural or just obvious, or if that distinction makes a difference as far as injectivity goes.

Thanks in advance.

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You are right, homomorphisms between free abelian groups are in general neither injective nor surjective. But here we have a bijection between the generators of the two groups - and this extends to a group isomorphism.

For any space $X$ let $\Sigma_k(X)$ denote the set of singular simplices $\sigma : \Delta^k \to X$. This is taken as the basis of the free abelian gropu $S_k(X)$.

The point is that each singular simplex $\sigma : \Delta^k \to X$ has a path-connected image and therefore $\sigma(\Delta^k)$ is contained in a unique path-component $X_\sigma$ of $X$. Let $\mathcal P$ denote the set of path components of $X$ and for each $P \in \mathcal P$ let $i_P : P \hookrightarrow X$ denote the inclusion map. Note that the $\Sigma_k(P)$ are pairwise disjoint. The function $$\psi : \bigcup_{P \in \mathcal P} \Sigma_k(P) \to \Sigma_k(X), \psi(\tau) = i_P \circ \tau$$ is clearly a bijection and $\bigcup_{P \in \mathcal P} \Sigma_k(P) $ is a basis of $\Sigma_{P \in \mathcal P} S_k(P)$.