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Let $X, Y$ be two topological spaces and $f:X\to Y$ a continuous map. Prove that the induced map $H_q(f):H_q(X)\to H_q(Y)$ is a monomorphism.

Idea: Let $[T_1],[T_2]\in H_q(X)$ s.t. $H_q(f)[T_1]=H_q(f)[T]$. This is equivalent that there exists $T'\in S_{q+1}Y$ with $$\partial_{q+1}^Y(T')=S_q(f)(T_1-T_2)$$ Also, the condition $[T_1]=[T_2]$ is equivalent to $\partial_{q+1}^X(T'')=(T_1-T_2)$ for some $T''\in S_{q+1}(X)$

From here, how can we prove the injectivity of $H_q(f)$?

Remember that $H_q(X)$ must be isomorphic to $\mathbb Z$

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    I've already shown the statement in italics is false, but did you mean to include the notion of path-connectness, from the title, in the theorem you asked about? – 2017-01-03
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    You're clearly leaving some details out. Why is $H_q(X)$ isomorphic to $\Bbb{Z}$? Is this some fixed $q$ or should it hold for all $q$? – 2017-01-03
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    Unfortunately I need to study more about singular homology. I made several mistakes in the previous statement. Thank you very much for your help! – 2017-01-03
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    @FUUNK1000: Since you agree that this question is mistaken, why not delete it? – 2017-01-04

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Consider $f:S^2 \to R^2 : (x, y, z) \mapsto (x, y)$, That's a continuous map of topological spaces, but since $H_2(S^2) = \Bbb Z$ and $H_2 (R^2) = 0$, $H_2(f)$ is not a monomorphism. The theorem you've asked us to prove is false.

By the way, what on earth does "Remember that $H_q(X)$ must be isomorphic to $\Bbb Z$" mean? You've mentioned this nowhere else!