# 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?

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.

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 $$-1$$).