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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.

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  • $\begingroup$ 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. $\endgroup$ – Michael Burr May 10 at 11:07
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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 $\mathbb Z$

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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".

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    $\begingroup$ well this is awkward... great minds think alike I guess? $\endgroup$ – Noel Lundström May 10 at 11:06

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