I am trying to make a picture in my head so I can understand and remember the rules.

So if $f \circ g$ is onto, it is onto because the function $f$ maps every element from a set $B$ to a set $C$ (thus $f$ is onto) and if $f \circ g$ is one-to-one then every element from set $A$ is mapping an element of set $B$ (and thus is one-to-one).

If both $f$ and $g$ is onto then $f \circ g$ is onto and if both $f$ and $g$ is one-to-one then $f \circ g$ is one-to-one and if both $f$ and $g$ are bijective then $f \circ g$ is bijective?

If $f \circ g$ is bijective, we can't say anything, but that $f$ is onto and that $g$ is one-to-one?

If $f$ is onto and $g$ is one-to-one, nothing can be said?

If $g$ is one-to-one and $g$ is onto, nothing can be said?

  • $\begingroup$ If $g$ is one-to-one and onto then $g$ is bijective. $\endgroup$ – Rankeya Nov 4 '12 at 19:58
  • $\begingroup$ was there any mistake? $\endgroup$ – George Milton Nov 4 '12 at 19:59
  • $\begingroup$ If you are asking if anything can be said about $g$ if it is one-to-one and onto, then what I am saying is that you can conclude the $g$ is bijective. I am not quite sure what mistake you are referring to. $\endgroup$ – Rankeya Nov 4 '12 at 20:00
  • $\begingroup$ If $g \circ f$ is bijective then really nothing more can be said beyond what you observe. For example: define $f: \{1,2\} \rightarrow \{1,2,3\}$ as $f(1) = 1, f(2) = 2$. Define $g: \{1,2,3\} \rightarrow \{1,2\}$ as $g(1) = 1, g(2) = 2, g(3) = 3$. Then $g \circ f$ is bijective. $\endgroup$ – Rankeya Nov 4 '12 at 20:07
  • $\begingroup$ i was just asking if i made any mistake in the post above $\endgroup$ – George Milton Nov 4 '12 at 20:21

This post intends to remove this question from the Unanswered list.

As noted in the comments, all of your assertions are correct.


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