# Is the induced map of a connected subspace on $0$th homology always injective?

I was asked whether or not the following is true:

Let $$Y$$ be a topological space, $$X \subset Y$$ a subspace, and $$f: X \hookrightarrow Y$$ the inclusion map. Then the induced map on homology $$f_{*}:H_n(X) \to H_n(Y)$$ is always injective.

This is of course false. Take $$S^1$$ embedded in $$S^2$$, par exemple. Then $$H_1(S^1) = \mathbb{Z}$$ while $$H_1(S^2) = 0$$.

But this got me thinking: What if we restrict ourselves to the $$0$$th homology?

If $$f: X \hookrightarrow Y$$ is the inclusion map, is the induced map $$H_0(X) \to H_0(Y)$$ on $$0$$th homology always injective?

Here a counterexample would be taking two distinct points $$x,y \in S^1$$. Then $$H_0(\{x,y\}) = \mathbb{Z} \oplus \mathbb{Z}$$ while $$H_0(S^1) = \mathbb{Z}$$.

But if we restrict ourselves even more:

If $$X$$ is a connected subspace of $$Y$$, and $$f: X \hookrightarrow Y$$ is the inclusion map, is the induced map $$H_0(X) \to H_0(Y)$$ on $$0$$th homology always injective?

I have been unable to find a counterexample, but am also unsure as to how one would prove this.

All help would be much appreciated.

• Since the $0$th homology counts the connected components, then yes this is true. It’s even easier to see in cohomology since the $0$th cohomology corresponds to locally constant functions. – Michael Burr May 10 at 11:07

This is not true, take the topologists sine curve embedded in $$\mathbb R^2$$ for example. It is connected but not path connected, it has two path components so its $$0-th$$ homology is $$\mathbb Z \oplus \mathbb Z$$ and of course the $$0-th$$ homology of $$\mathbb R^2$$ is $$\mathbb Z$$
No, you need path-connectedness. Consider the inclusion of $$X$$ in $$Y=\Bbb R^2$$ where $$X$$ is a connected but not path-connected space, for instance the closure of the "topologist's sine curve".