$f_*\colon H_0(X)\to H_0(Y)~\text{is an isomorphism}~\Leftrightarrow ~f_*\colon \tilde{H}_0(X)\to \tilde{H}_0(Y)~\text{is an isomorphism}$ For a continuous map $f\colon X\to Y$, does the following hold? $$ f_*\colon H_0(X)\to H_0(Y)~\text{is an isomorphism}~\Leftrightarrow ~f_*\colon \tilde{H}_0(X)\to \tilde{H}_0(Y)~\text{is an isomorphism}$$
I know that the former map is an isomorphism provided that $f(X)$ intersects each path component of $Y$.
I think this must be true, but I have no idea to prove it.
Thanks in advance.
 A: Recall that reduced homology is defined for pointed spaces and $\tilde{H}_n(X) = H_n(X, x_0)$ where $x_0$ is the basepoint. We have a long exact sequence for the pair $(X, x_0)$ which at the bottom takes the form
$$ \dots \to H_0(x_0) \cong G \to H_0(X) \to H_0(X, x_0)$$
Moreover, the map $H_0(x_0) \to H_0(X)$ splits via the map induced by $X \to x_0$, so it's actually a split-exact sequence 
$$G \to H_0(X) \cong (G \oplus H_0(X, x_0)) \to H_0(X, x_0) $$
and this splitting is natural wrt pointed maps. Now, remember that the long exact sequence of the pair is also natural, so given the pointed map $f\colon (X, x_0) \to (Y, y_0)$ we get a commutative diagram
$\require{AMScd}$
\begin{CD}
G @>>> G\oplus\tilde{H}_0(X)\\
@V{f_*}VV @V{f_*}VV\\
G @>>> G\oplus\tilde{H}_0(Y)\\
\end{CD}
The map on the left is always an isomorphism, and by naturality, the map on the right restricts to this same isomorphism on the $G$ factor. It follows that the map on the right is an isomorphism iff it restricts to an isomrphism on reduced homology.

Remark: You can also prove this just by counting components, but I wanted to highlight that this is a general phenomenon: if $h_*$ is any generalized homology theory then there is a natural splitting $h_n(X) \cong \tilde{h}_n(X) \oplus h_n(x_0)$ for any pointed space $(X, x_0)$, so the statement you want holds for all homology theories and all degrees.
