I am wondering whether $\mathbb{Z}\times\mathbb{Z}$ is not injective to $\mathbb{Z}$. Similarly, whether $\mathbb{Z}*\mathbb{Z}$ is not injective to $\mathbb{Z}$.

I got it from the Hatcher's Algebraic Topology question in Section 1.1 exercise 16 that asks to show there are no restractions $r:S^1\times D^2\to S^1\times S^1$. We have $\pi_1(S^1\times D^2)\simeq\mathbb{Z}$ and $\pi_1(S^1\times S^1)\simeq\mathbb{Z}\times\mathbb{Z}$. I am trying to reach the conclusion that $\mathbb{Z}\times\mathbb{Z}$ is not injective to $\mathbb{Z}$ since a space retracts to its subspace has their induced homomorphim injective, in that way I can get a contradiction. But is there really no injection between $\mathbb{Z}\times\mathbb{Z}$ and $\mathbb{Z}$?

As far as I know, cartesian product of countable set is countable, so why could there be no injection between $\mathbb{Z}\times\mathbb{Z}$ and $\mathbb{Z}$?

Also, I am not sure whether $\mathbb{Z}*\mathbb{Z}$ is countably or uncountably infinite, if it is countably infinite, why could there also be no injection?

Does anybody has any good explanation on this? Any hints would also be appreciated. Thanks.

  • $\begingroup$ There are injections between $\mathbb Z$ and $\mathbb Z^2$. $\endgroup$ – Git Gud Oct 14 '16 at 13:17
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    $\begingroup$ What is $\mathbb Z*\mathbb Z$? $\endgroup$ – Git Gud Oct 14 '16 at 13:21
  • $\begingroup$ Maybe you are referring to a group structure on those sets? $\endgroup$ – nombre Oct 14 '16 at 13:21
  • $\begingroup$ @GitGud It is the free product (coproduct in groups) of $\mathbb {Z}$'s. Also known as the free group on two generators. $\endgroup$ – Paul Plummer Oct 14 '16 at 13:29

There are injections, like you said they are countable sets, so there are even bijections between them.

There is no injective group homomorphism though. This comes from the fact there is no subgroup of $\mathbb{Z}$ which is isomorphic to $\mathbb{Z} \times \mathbb{Z}$ or $\mathbb{Z}*\mathbb{Z}$. I am guessing that is what your sources were referring to (context is important).

If $H$ is isomorphic to a subgroup $K$ of $G$, then there exists an isomorphism $\phi: H \to K$, and there is a natural injection $i :K \to G$, $i \circ \phi:H \to G$ is an injection. If there is an injective homomorphism $\phi:H \to G$, then $\phi(H) \cong H$, which if you want you can view as some sort of application of the first isomorphism theorem, since the kernel of the homomorphism is trivial, although it is easier than that.

  • $\begingroup$ Yes you're right. Thanks. Do you mind stating the statement of the fact that you stated? Is it that if a subgroup of a group $G$ is isomorphic to another group $H$ then there is an injective homomorphism between $G$ and $H$? Is it related to the first isomorphism theorem in any way? $\endgroup$ – user338393 Oct 14 '16 at 13:31
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    $\begingroup$ "Do you mind stating the statement of the fact that you stated?"... $\endgroup$ – Paul Plummer Oct 14 '16 at 13:35
  • $\begingroup$ Sorry for the question, maybe because I have never seen such statement before. I am just wondering whether it is related to the first isomorphism theorem? $\endgroup$ – user338393 Oct 14 '16 at 13:38
  • $\begingroup$ @user338393 I understood what you were wanting, I just thought that sentence was funny. I edited in a quick explanation. Although you should try to prove it yourself, it is good exercise to make sure you are comfortable with the basic algebra. $\endgroup$ – Paul Plummer Oct 14 '16 at 13:42
  • $\begingroup$ Yes, thanks so much. $\endgroup$ – user338393 Oct 14 '16 at 13:45

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