If one has two path-connected spaces, must a map induced by $f:X\to Y$ be an isomorphism on $H_0(X)\to H_0(Y)$? Say $X$ and $Y$ are two $n$-dimensional, connected CW complexes, (so they are path connected), and say $f:X\to Y$ is a continuous map. Must it be true that $f_*:H_0(X)\to H_0(Y)$ is an isomorphism? I know that $H_0(X)\cong H_0(Y)\cong \Bbb{Z}$, but I don't know that $f_*:H_0(X)\to H_0(Y)$ will induce such an isomorphism.
This seems to be somewhat similar to this question, so I'm inclined to say yes.
Part of the reason I am unsure about this is because I don't totally understand what each element in $H_0(X)$ represents. I know that overall, $H_0(X)$ counts the number of path components of $X$, but what, for example, would the element isomorphic to $2$ in $H_0(X)$ represent?
 A: The elements of $H_0(X)$ are in general formal sums of classes of points $x\in X$ where two points are equivalent if they are in the same path component.
As we have only one path component $H_0(X) = \mathbf Z \cdot [x]$ where $[x]$ is the class of any point $x$ we pick, (as $[x] = [x']$ for any other point $x'\in X$).
The pushforward map on homology $f_*\colon H_0(X) \to H_0(Y)$ induced by $f\colon X \to Y$ simply sends each $[x] \in H_0(X)$ to $[f(x)] \in H_0(Y)$, so if $H_0(X) = \mathbf Z \cdot [x]$ and $H_0(Y) = \mathbf Z \cdot [y]$ then
$$
f_*[x] = [f(x)] = [y]
$$
by definition and equivalence as $f(x)$ is in the same path component as $y$.
So this map could be called the identity if we identify both sides with $\mathbf Z$ sending the generator to the class of a point, but personally I think it is better to always write $\mathbf Z\cdot [x]$ to remind ourselves what the class represents.
If you identify $\mathbf Z \cong \mathbf Z \cdot [x]$ then the element 2 could be said to represent $2$ copies of the point $x$ I suppose, or really two copies of the whole component as all points in a component are equivalent so there is no reason to prefer one over the other.
A: Nice question. By definition of singular homology, the group $H_0(X)$ is the quotient of the free abelian group on maps $\{pt\} \to X$ by the subgroup of those formal sums of the form $\phi(1) - \phi(0)$ where $\phi: [0, 1] \to X$ is a path. We can think of a map $\{pt\} \to X$ as just a point on $X$, of course.
We have the degree map 
$$\deg: H_0(X) \to \mathbb{Z}
$$ which just takes a formal sum of maps $\{pt\} \to X$ to the sum of their coefficients. This map is clearly onto, and well defined mod the equivalence relation (which only identifies degree $0$ formal sums to $0$). If $X$ is path-connected, then this map is also injective, since any degree zero sum  is generated by formal sums of the form $(P - Q)$, and a path connecting $P$ to $Q$ exhibits these formal sums as zero in $H_0(X)$.
Now we see that the map $X \to Y$ must be an isomorphism, and even map the canonical generator to the canonical generator (in other words, a priori there are two abelian group isomorphisms from $\mathbb{Z} \to \mathbb{Z}$, but the identifications of $H_0(X)$ with $\mathbb{Z}$ and $H_0(Y)$ with $\mathbb{Z}$ are canonical, and the map $\mathbb{Z} \to \mathbb{Z}$ we get is the identity, not $-1$).
