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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.