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Check that the map $F:X\times Y\rightarrow Y \times X$, defined by $(x,y) \rightarrowtail (y,x)$ is bijective. Describe the relationship between mutually inverse maps $f:X \rightarrow Y$ and $f^{-1}:Y \rightarrow X$.

Question is from book "Mathematical Analysis" by Zorich, chapter 1.3 - Function, question 5d.

My attempt at a proof: First, let $f:X \rightarrow Y$ and $g:Y \rightarrow X$.

Then, to show that f is injective, we note that $g \circ f = id_X$, which implies that g is the left inverse of f, hence f is injective.

To show that f is surjective, we note that $f \circ g = id_Y$. So g is the right inverse of f, hence f is surjective. Hence f is bijective and g = $f^{-1}$, which is bijective too. Composition of bijective functions is bijective, so $g \circ f = F:X\times Y\rightarrow Y \times X$ is bijective.

Second part of the question is a bit unclear to me though. The relationship between graphs of f and $f^{-1}$ is obviously, as it is stated in the actual question, is that they are mutually inverse functions. However, I am not sure what else I can say regarding that. The definition of a graph of function is {$(x,y) \in X \times Y | y = f(x)$}. So for the inverse it would be {$(y,x) \in Y \times X | x = f(y)$}. But just stating the definitions doesn't seem like a sufficient answer.

Would greatly appreciate any corrections in both logic and notation. In particular, I am very not sure that I'm allowed to say that $g \circ f = id_X$ in the function $f:X \rightarrow Y$, as I defined it, which is what my whole argument is based on.

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    $\begingroup$ Although it is obvious, you should define $g$ in your proof. $\endgroup$
    – DanDan面
    Commented Jun 26, 2023 at 23:20
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    $\begingroup$ Also, you seem to prove that $f$ is bijective, rather than $F$. $\endgroup$
    – DanDan面
    Commented Jun 26, 2023 at 23:22
  • $\begingroup$ @DanDan0101 But if f is bijective, then g, its inverse, is also bijective? So their composition is bijective. $\endgroup$
    – John Doe
    Commented Jun 27, 2023 at 10:13
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    $\begingroup$ @JohnDoe But $F$ is not their composition, as I said. We don't even know what "they" are $\endgroup$
    – FShrike
    Commented Jun 27, 2023 at 10:23

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This doesn't make too much sense. You haven't defined $g$! Or $f$. You claim $F=g\circ f$ but if $f:X\to Y$ and $g:Y\to X$, then $g\circ f:X\to X$ whereas $F:X\times Y\to Y\times X$. So there is a serious domain and codomain error (also, you claim $g\circ f=\mathrm{id}$, whatever $g,f$ are... and $F$ is not the identity).

For the second part, try to relate the graphs of $f,f^{-1}$ with the function $F$.

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  • $\begingroup$ Thanks for the reply! Would you give me a hint on what I should start with to rewrite my proof? And if any parts of my existing proof could be used? $\endgroup$
    – John Doe
    Commented Jun 27, 2023 at 10:06
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    $\begingroup$ @JohnDoe With respect, I don't think any of what you wrote can be used: it is founded on these undefined "$f,g$". You want to show $F$ is bijective. Can you write down an inverse to $F$? Hopefully. Or, can you check that: $F$ injects and: $F$ surjects? Try to do direct checks. Mistakes newcomers sometimes make is get lost in repeating definitions, which you seemed to do in this case $\endgroup$
    – FShrike
    Commented Jun 27, 2023 at 10:13
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    $\begingroup$ @JohnDoe Yeah that's right. $(y,x)\mapsto(x,y)$ is an inverse of $F$. $\endgroup$
    – FShrike
    Commented Jun 29, 2023 at 14:38
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    $\begingroup$ @JohnDoe (1) Yes, to show $F$ has an inverse suffices to show it is injective, and also surjective, so bijective. (2) Yes, as long as you actually check the composite is the identity. (3) The connection is that $F(\text{something})=\text{something else}$ where the 'something's are related to the graphs $\endgroup$
    – FShrike
    Commented Jun 29, 2023 at 15:50
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    $\begingroup$ Ok, thanks again for you time! $\endgroup$
    – John Doe
    Commented Jun 29, 2023 at 15:55

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