Mixing
What does it mean that there is an “isomorphism of homsets” due to an exponential object?
$begingroup$
$X^Y$ together with a morphism $bf{ apply}$ $:X^Ytimes Yto X$ is an exponential object of $X$ and $Y$ if for each $Z$ and each morphism $f:Ztimes Y to X$, there is a unique morphism $lambda f:Zto X^Y$ such that $lambda f times id_Y circ bf {apply}$ $ = f$.
It is obvious that for every $f$ there is a $lambda f$. But two things are not obvious to me:
Is there necessarily for every morphism $lambda g:Zto X^Y$ a corresponding morphism $g:Ztimes Y to X$? Couldn't we have a category where there are these $lambda f:Zto X^Y$, but also additional extraneous morphisms $lambda g:Zto X^Y$ that have no corresponding $g$? Can't we always add such extraneous morphisms?
Assuming that the previous question is clarified, and there is indeed a bijection between these homsets, why does wikipedia call it a "isomorphism" of homsets? An isomorphism is only an isomorphism within a category, and I don't know what category they're talking about.
category-theory
asked Jan 29 at 8:35
user56834user56834
3,37921253
$endgroup$
add a comment |
$begingroup$
$X^Y$ together with a morphism $bf{ apply}$ $:X^Ytimes Yto X$ is an exponential object of $X$ and $Y$ if for each $Z$ and each morphism $f:Ztimes Y to X$, there is a unique morphism $lambda f:Zto X^Y$ such that $lambda f times id_Y circ bf {apply}$ $ = f$.
It is obvious that for every $f$ there is a $lambda f$. But two things are not obvious to me:
Is there necessarily for every morphism $lambda g:Zto X^Y$ a corresponding morphism $g:Ztimes Y to X$? Couldn't we have a category where there are these $lambda f:Zto X^Y$, but also additional extraneous morphisms $lambda g:Zto X^Y$ that have no corresponding $g$? Can't we always add such extraneous morphisms?
Assuming that the previous question is clarified, and there is indeed a bijection between these homsets, why does wikipedia call it a "isomorphism" of homsets? An isomorphism is only an isomorphism within a category, and I don't know what category they're talking about.
category-theory
asked Jan 29 at 8:35
user56834user56834
3,37921253
$endgroup$
4
$begingroup$
I don't mean to offend, but you've asked a lot of questions tonight which would be covered by any introductory text on the subject. Tom Leinster's Basic Category Theory has a freely downloadable version available on the Arxiv, and you might benefit from actually spending some time with it. As it stands, it feels like you're skimming Wikipedia and then asking us to do the work of a textbook.
$endgroup$
– Malice Vidrine
Jan 29 at 9:03
1
$begingroup$
@MaliceVidrine, I am reading a textbook, and following lectures. I just find some of it hard enough to follow that I'm confused.
$endgroup$
– user56834
Jan 29 at 12:13
add a comment |
$begingroup$
$X^Y$ together with a morphism $bf{ apply}$ $:X^Ytimes Yto X$ is an exponential object of $X$ and $Y$ if for each $Z$ and each morphism $f:Ztimes Y to X$, there is a unique morphism $lambda f:Zto X^Y$ such that $lambda f times id_Y circ bf {apply}$ $ = f$.
It is obvious that for every $f$ there is a $lambda f$. But two things are not obvious to me:
Is there necessarily for every morphism $lambda g:Zto X^Y$ a corresponding morphism $g:Ztimes Y to X$? Couldn't we have a category where there are these $lambda f:Zto X^Y$, but also additional extraneous morphisms $lambda g:Zto X^Y$ that have no corresponding $g$? Can't we always add such extraneous morphisms?
Assuming that the previous question is clarified, and there is indeed a bijection between these homsets, why does wikipedia call it a "isomorphism" of homsets? An isomorphism is only an isomorphism within a category, and I don't know what category they're talking about.
category-theory
asked Jan 29 at 8:35
user56834user56834
3,37921253
$endgroup$
$X^Y$ together with a morphism $bf{ apply}$ $:X^Ytimes Yto X$ is an exponential object of $X$ and $Y$ if for each $Z$ and each morphism $f:Ztimes Y to X$, there is a unique morphism $lambda f:Zto X^Y$ such that $lambda f times id_Y circ bf {apply}$ $ = f$.
It is obvious that for every $f$ there is a $lambda f$. But two things are not obvious to me:
Is there necessarily for every morphism $lambda g:Zto X^Y$ a corresponding morphism $g:Ztimes Y to X$? Couldn't we have a category where there are these $lambda f:Zto X^Y$, but also additional extraneous morphisms $lambda g:Zto X^Y$ that have no corresponding $g$? Can't we always add such extraneous morphisms?
Assuming that the previous question is clarified, and there is indeed a bijection between these homsets, why does wikipedia call it a "isomorphism" of homsets? An isomorphism is only an isomorphism within a category, and I don't know what category they're talking about.
category-theory
category-theory
asked Jan 29 at 8:35
user56834user56834
3,37921253
asked Jan 29 at 8:35
user56834user56834
3,37921253
asked Jan 29 at 8:35
user56834user56834
3,37921253
asked Jan 29 at 8:35
user56834user56834
3,37921253
asked Jan 29 at 8:35
user56834user56834
3,37921253
3,37921253
4
$begingroup$
I don't mean to offend, but you've asked a lot of questions tonight which would be covered by any introductory text on the subject. Tom Leinster's Basic Category Theory has a freely downloadable version available on the Arxiv, and you might benefit from actually spending some time with it. As it stands, it feels like you're skimming Wikipedia and then asking us to do the work of a textbook.
$endgroup$
– Malice Vidrine
Jan 29 at 9:03
1
$begingroup$
@MaliceVidrine, I am reading a textbook, and following lectures. I just find some of it hard enough to follow that I'm confused.
$endgroup$
– user56834
Jan 29 at 12:13
add a comment |
4
$begingroup$
I don't mean to offend, but you've asked a lot of questions tonight which would be covered by any introductory text on the subject. Tom Leinster's Basic Category Theory has a freely downloadable version available on the Arxiv, and you might benefit from actually spending some time with it. As it stands, it feels like you're skimming Wikipedia and then asking us to do the work of a textbook.
$endgroup$
– Malice Vidrine
Jan 29 at 9:03
1
$begingroup$
@MaliceVidrine, I am reading a textbook, and following lectures. I just find some of it hard enough to follow that I'm confused.
$endgroup$
– user56834
Jan 29 at 12:13
4
4
$begingroup$
I don't mean to offend, but you've asked a lot of questions tonight which would be covered by any introductory text on the subject. Tom Leinster's Basic Category Theory has a freely downloadable version available on the Arxiv, and you might benefit from actually spending some time with it. As it stands, it feels like you're skimming Wikipedia and then asking us to do the work of a textbook.
$endgroup$
– Malice Vidrine
Jan 29 at 9:03
$begingroup$
I don't mean to offend, but you've asked a lot of questions tonight which would be covered by any introductory text on the subject. Tom Leinster's Basic Category Theory has a freely downloadable version available on the Arxiv, and you might benefit from actually spending some time with it. As it stands, it feels like you're skimming Wikipedia and then asking us to do the work of a textbook.
$endgroup$
– Malice Vidrine
Jan 29 at 9:03
1
1
$begingroup$
@MaliceVidrine, I am reading a textbook, and following lectures. I just find some of it hard enough to follow that I'm confused.
$endgroup$
– user56834
Jan 29 at 12:13
$begingroup$
@MaliceVidrine, I am reading a textbook, and following lectures. I just find some of it hard enough to follow that I'm confused.
$endgroup$
– user56834
Jan 29 at 12:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(i) : No : start from $h : Zto X^Y$; then you have $htimes id_Y : Ztimes Yto X^Ytimes Y$, and then you can compose with $mathbf{apply}$ to get $g:= mathbf{apply}circ (htimes id_Y) : Ztimes Yto X$.
Then by uniqueness, $lambda g = h$.
(ii) : A bijection between sets is an isomorphism in the category of sets $mathbf{Set}$.
But here it's even lore than that, because the isomorphism $hom(Ztimes Y, X) cong hom(Z, X^Y)$ is natural in all three variables, so it is actually an isomorphism between two functors in the category of functors $C^{op}times C^{op}times Cto mathbf{Set}$
answered Jan 29 at 9:00
MaxMax
15.8k11143
$endgroup$
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1 Answer
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$begingroup$
(i) : No : start from $h : Zto X^Y$; then you have $htimes id_Y : Ztimes Yto X^Ytimes Y$, and then you can compose with $mathbf{apply}$ to get $g:= mathbf{apply}circ (htimes id_Y) : Ztimes Yto X$.
Then by uniqueness, $lambda g = h$.
(ii) : A bijection between sets is an isomorphism in the category of sets $mathbf{Set}$.
But here it's even lore than that, because the isomorphism $hom(Ztimes Y, X) cong hom(Z, X^Y)$ is natural in all three variables, so it is actually an isomorphism between two functors in the category of functors $C^{op}times C^{op}times Cto mathbf{Set}$
answered Jan 29 at 9:00
MaxMax
15.8k11143
$endgroup$
add a comment |
$begingroup$
(i) : No : start from $h : Zto X^Y$; then you have $htimes id_Y : Ztimes Yto X^Ytimes Y$, and then you can compose with $mathbf{apply}$ to get $g:= mathbf{apply}circ (htimes id_Y) : Ztimes Yto X$.
Then by uniqueness, $lambda g = h$.
(ii) : A bijection between sets is an isomorphism in the category of sets $mathbf{Set}$.
But here it's even lore than that, because the isomorphism $hom(Ztimes Y, X) cong hom(Z, X^Y)$ is natural in all three variables, so it is actually an isomorphism between two functors in the category of functors $C^{op}times C^{op}times Cto mathbf{Set}$
answered Jan 29 at 9:00
MaxMax
15.8k11143
$endgroup$
add a comment |
$begingroup$
(i) : No : start from $h : Zto X^Y$; then you have $htimes id_Y : Ztimes Yto X^Ytimes Y$, and then you can compose with $mathbf{apply}$ to get $g:= mathbf{apply}circ (htimes id_Y) : Ztimes Yto X$.
Then by uniqueness, $lambda g = h$.
(ii) : A bijection between sets is an isomorphism in the category of sets $mathbf{Set}$.
But here it's even lore than that, because the isomorphism $hom(Ztimes Y, X) cong hom(Z, X^Y)$ is natural in all three variables, so it is actually an isomorphism between two functors in the category of functors $C^{op}times C^{op}times Cto mathbf{Set}$
answered Jan 29 at 9:00
MaxMax
15.8k11143
$endgroup$
(i) : No : start from $h : Zto X^Y$; then you have $htimes id_Y : Ztimes Yto X^Ytimes Y$, and then you can compose with $mathbf{apply}$ to get $g:= mathbf{apply}circ (htimes id_Y) : Ztimes Yto X$.
Then by uniqueness, $lambda g = h$.
(ii) : A bijection between sets is an isomorphism in the category of sets $mathbf{Set}$.
But here it's even lore than that, because the isomorphism $hom(Ztimes Y, X) cong hom(Z, X^Y)$ is natural in all three variables, so it is actually an isomorphism between two functors in the category of functors $C^{op}times C^{op}times Cto mathbf{Set}$
answered Jan 29 at 9:00
MaxMax
15.8k11143
answered Jan 29 at 9:00
MaxMax
15.8k11143
answered Jan 29 at 9:00
MaxMax
15.8k11143
answered Jan 29 at 9:00
MaxMax
15.8k11143
15.8k11143
add a comment |
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$begingroup$
I don't mean to offend, but you've asked a lot of questions tonight which would be covered by any introductory text on the subject. Tom Leinster's Basic Category Theory has a freely downloadable version available on the Arxiv, and you might benefit from actually spending some time with it. As it stands, it feels like you're skimming Wikipedia and then asking us to do the work of a textbook.
$endgroup$
– Malice Vidrine
Jan 29 at 9:03
1
$begingroup$
@MaliceVidrine, I am reading a textbook, and following lectures. I just find some of it hard enough to follow that I'm confused.
$endgroup$
– user56834
Jan 29 at 12:13