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I was wondering if the singular homology functor preserve injectivity and surjectivity? I've been trying to figure out a proof or counterexample for ages now and I just can't.

This came up when I was looking at the reduced homology $H_p(S^{n},S^{n-1})$. To calculate it, I have looked at the canonical injection $$\iota: S^{n-1} \longrightarrow S^{n}.$$ I'm viewing $S^{n-1}$ as the equator of $S^n$, in this case does the functor preserve injectivity? That is, I want to say that $\iota_*$ is injective. Is this true? Thanksarinos

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    $\begingroup$ It's false, there is an injection $i : \mathbb S^1 \to \mathbb R^2$ but the induced map is of course not injective. For the surjectivity, think to the quotient map $ q : [0,1] \to \mathbb S^1$ which glue 0 and 1 together. $\endgroup$ – user171326 May 12 '15 at 9:27
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    $\begingroup$ Based on your comments below, I have to ask: what is $\iota_*$? Apparently it's not just $\iota_* : H_*(S^{n-1}) \to H_*(S^n)$? $\endgroup$ – Najib Idrissi May 12 '15 at 14:44
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I don't understand what you mean by relative homology preserving injectivity and surjectivity, so consider instead the following general point.

No interesting homotopy-invariant functor on spaces can preserve injections or surjections. The reason is that every map is homotopy equivalent to an injection and also to a surjection! There are very nice explicit constructions accomplishing this called the mapping cylinder and mapping cocylinder respectively.

In other words, "surjection" and "injection" aren't homotopy-theoretically meaningful concepts.

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  • $\begingroup$ A Q&A related to your point. $\endgroup$ – Najib Idrissi May 13 '15 at 14:18
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Your example already answers your question. $\iota_*: H_*(S^{n-1})\to H_*(S^n)$ is not injective. Also, $\pi: S^2\to\mathbb{RP}^2$ is surjective but the induced map on homology is not.

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  • $\begingroup$ Why is it not injective? $\endgroup$ – AlgebraicTopologyfourlyfffffff May 12 '15 at 9:31
  • $\begingroup$ $H_{n-1}(S^{n-1})\cong\mathbb{Z}$ but $H_{n-1}(S^n)\cong 0$. $\endgroup$ – Alex Fok May 12 '15 at 9:33
  • $\begingroup$ We're talking about the relative homology. sorry I didn't make that clear. English isn't my second language. $\endgroup$ – AlgebraicTopologyfourlyfffffff May 12 '15 at 9:47

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