Maps induced on (co)homology and change of coefficients Suppose I have a map $f: X \rightarrow Y$ such that the kernel and cokernel of $f^*: H^n(Y; \mathbb{Z}) \rightarrow H^n(X; \mathbb{Z})$ are both finite abelian groups.
Is is true that the map $f^*: H^n(Y; \mathbb{Q}) \rightarrow H^n(X; \mathbb{Q})$ is surjective? If so, what is the best way to show this? Can anybody please recommend a good reference as to how induced maps behave under different coefficients?
 A: If the integral cohomology of $X,Y$ is finitely generated then $f_\mathbb{Q}^\ast: H^n(X;\mathbb{Q}) \to H^n(Y;\mathbb{Q})$ is even an isomorphism. 
Proof: By the Universal Coefficient Theorem (UTC) [Spanier, Chap. 5, sec. 5, Thorem 10] there is a short exact sequence 
$$0 \to H^i(X) \otimes \mathbb{Q} \to H^i(X;\mathbb{Q}) \to Tor_1^\mathbb{Z}(H^{i+1}(X),\mathbb{Q}) \to 0$$
and since $\mathbb{Q}$ is a flat $\mathbb{Z}$-module, the $Tor$-term vanishes. So by naturality of UCT there is a commutative diagramm 
$$\begin{array}{ccc}
H^i(X) \otimes \mathbb{Q} & \cong & H^i(X;\mathbb{Q}) \newline 
{\scriptstyle f_\mathbb{Z}^\ast \otimes id}\downarrow & & \downarrow {\scriptstyle  f_\mathbb{Q}^\ast}\newline 
H^i(Y) \otimes \mathbb{Q} & \cong & H^i(Y;\mathbb{Q}) 
\end{array}$$ 
Hence it suffices to show that the left arrow is an isomorphism. To simplify notation set $A := H^i(X),\;B:= H^i(Y),\; h := f_\mathbb{Z}^\ast$. Tensoring the exact sequence 
$$0 \to \ker(h) \to A \xrightarrow{h} B \to B/\text{im}(h) \to 0$$
with the flat module $\mathbb{Q}$ and using that $ker(h), B/\text{im}(h)$ are finite yields the exact sequence 
$$0 \to 0 \to A \otimes \mathbb{Q} \xrightarrow{h \otimes id} B \otimes \mathbb{Q} \to 0 \to 0$$
i.e. $h \otimes id$ is an isomorphism. QED
