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Not sure if this question has been asked before, but anyways. I was wondering if someone wouldn't mind clarifying what the question is asking:

"Among the examples above, the inclusion maps are one-to-one, but, except in some trivial special cases, the projection are not. Exercise: What special cases?" -Halmos Section 8 Pg. 32

Just as background, he presents the the projection functions onto $X$ and $Y$ from $X \times Y$, and the canonical map; sorry, it is really a lot to write, but there is an online copy pg.32.

What I am most confused about is the inclusion map for the (seemly ambiguous) "projection". What I am thinking is $i: P \times Q \rightarrow X \times Y$, $P \subset X$ and $Q \subset Y$. However, the only way I could see this failing 1-1 is in cases $P \times Q = \emptyset$, yet I am not sure if that is what this question is asking.

EDIT: One of my biggest questions is if $i$ as stated above is indeed the "projection" inclusion map. It seems like it is not the projection.

Any help would be appreciated.

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  • $\begingroup$ The function with an empty domain is the empty function, and it is 1-1, vacuously. (There are no two inputs with the same output!) $\endgroup$ – Alan Nov 13 '15 at 5:36
  • $\begingroup$ @Alan You're totally right, those were just my thoughts. (not good ones) $\endgroup$ – 9301293 Nov 13 '15 at 5:37
  • $\begingroup$ Yeah, I'm playing around with trying to find a special case where it fails to be injective myself $\endgroup$ – Alan Nov 13 '15 at 5:37
  • $\begingroup$ There are plenty of cases where it fails to be 1-1, e.g. $X=Y=\{0,1\}$. $\endgroup$ – BrianO Nov 13 '15 at 5:40
  • $\begingroup$ Oh, doh, misparsed it. The inclusions are always injective. He's saying the projections are almost never injective $\endgroup$ – Alan Nov 13 '15 at 5:44
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In order of increasing triviality:

  • If all the factors of a Cartesian product are nonempty and all except one are one-element sets, the projection to the factor with more than one element is 1-1, but the others are not.
  • If all factors of a Cartesian product are one-element sets, then all projections are 1-1.
  • If any of the factors are $\emptyset$, the projections are 1-1 (vacuously: the product is then $\emptyset$).
  • All projections from the Cartesian product of an empty collection are 1-1. (Vacuously: the product is $\{\emptyset\}$, but as there are no factors, there are no projections, so "all of them" are 1-1, as well as not 1-1, and equal to 17.)

These are the only cases where (some or all of) the projections are 1-1.

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  • $\begingroup$ Just to clarify, then, Halmos was asking about $i:P×Q→X×Y$, $P⊂X$, and $Q⊂Y$? $\endgroup$ – 9301293 Nov 13 '15 at 5:41
  • $\begingroup$ No, he's talking about $\pi_X\colon (x,y)\mapsto x\colon X\times Y\to X$ and $\pi_Y\colon (x,y)\mapsto y\colon X\times Y\to Y$. $\endgroup$ – BrianO Nov 13 '15 at 5:43
  • $\begingroup$ In the exceptional case, only the projection to the only coordinate with several elements would be 1-1. $\endgroup$ – dafinguzman Nov 13 '15 at 5:43
  • $\begingroup$ @dafinguzman Yes that's so. I'll clarify that. $\endgroup$ – BrianO Nov 13 '15 at 5:44
  • $\begingroup$ By how is that an inclusion map?: Maybe this is where I went wrong.... $\endgroup$ – 9301293 Nov 13 '15 at 5:44

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