Map between pairs $(X,A) \to (Y,B)$ induces homomorphism taking $C_n(A)$ to $C_n(B)$

Question

I'm going through relative homology in Hatcher's Algebraic Topology.

Just as there are induced homomorphisms for nonrelative homology, a map $f: (X,A) \to (Y,B)$ — i.e., such that $f(A) \subset B$ — induces homomorphisms $f_\sharp: C_n(X,A) \to C_n(Y,B)$. Of course, for this to work, $f_\sharp: C_n(X) \to C_n(Y)$ must "take $C_n(A)$ to $C_n(B)$" (Hatcher, pg. 118).

However, this is not immediately obvious to me. Since $f$ restricted to $A \to B$ can fail to be surjective, why should every $n$-chain in $B$ be mapped to by an $n$-chain in $A$? In other words, isn't it possible that there exists a chain in $B$ which is not mapped to by a chain in $A$ via $f_\sharp$?

Answer 1

Nevermind, it appears that it was not meant that $f_\sharp$ takes $C_n(A)$ (on)to $C_n(B)$, and its restriction to $C_n(A) \to C_n(B)$ is, indeed, not surjective in general.

I think a proof that maps between pairs $(X,A) \to (Y,B)$ induces a well-defined homomorphism $f_\sharp: C_n(X,A) \to C_n(Y,B)$ goes like this:

Suppose $[c] \in C_n(X,A)$. Given $f_\sharp: C_n(X) \to C_n(Y)$, define $f_\sharp: C_n(X,A) \to C_n(Y,B)$ by $f_\sharp([c]) = [f_\sharp(c)] \in C_n(Y,B)$.

To see that this is well-defined, let $a \in C_n(A)$ and $b \in C_n(B)$. Then \begin{align} f_\sharp([c+a]) &= [f_\sharp(c+a)] \\ &= [f_\sharp(c)+f_\sharp(a)] \\ &= [f_\sharp(c)+b] \\ &= [f_\sharp(c)] \\ &= f_\sharp([c]). \end{align}

Thus, the surjectivity of $f_\sharp: C_n(A) \to C_n(B)$ was never needed.