Mixing
are two different pairs of objects allowed to share the same morphism in a category?
$begingroup$
Normally I think of a morphism as belonging to two objects. e.g. for objects $a,b$ I think of a morphism $f$ between $a$ and $b$ as having the "type" $f:ato b$.
But I can't see in the definition of category that a morphism has to be connected specifically to two objects.
Can we have a category with distinct objects $a,b,c,d$ where $Hom(a,b)cap Hom(c,d)neq emptyset$?
example: we could have the morphisms be distances between objects where the objects are elements of a metric space. Then $Hom(a,b)= Hom(c,d)$ if $dist(a,b)=dist(c,d)$.
category-theory
asked Jan 28 at 7:13
user56834user56834
3,35921253
$endgroup$
add a comment |
$begingroup$
Normally I think of a morphism as belonging to two objects. e.g. for objects $a,b$ I think of a morphism $f$ between $a$ and $b$ as having the "type" $f:ato b$.
But I can't see in the definition of category that a morphism has to be connected specifically to two objects.
Can we have a category with distinct objects $a,b,c,d$ where $Hom(a,b)cap Hom(c,d)neq emptyset$?
example: we could have the morphisms be distances between objects where the objects are elements of a metric space. Then $Hom(a,b)= Hom(c,d)$ if $dist(a,b)=dist(c,d)$.
category-theory
asked Jan 28 at 7:13
user56834user56834
3,35921253
$endgroup$
$begingroup$
I like it when arrows in a category have well defined sources and targets....
$endgroup$
– Lord Shark the Unknown
Jan 28 at 7:24
add a comment |
$begingroup$
Normally I think of a morphism as belonging to two objects. e.g. for objects $a,b$ I think of a morphism $f$ between $a$ and $b$ as having the "type" $f:ato b$.
But I can't see in the definition of category that a morphism has to be connected specifically to two objects.
Can we have a category with distinct objects $a,b,c,d$ where $Hom(a,b)cap Hom(c,d)neq emptyset$?
example: we could have the morphisms be distances between objects where the objects are elements of a metric space. Then $Hom(a,b)= Hom(c,d)$ if $dist(a,b)=dist(c,d)$.
category-theory
asked Jan 28 at 7:13
user56834user56834
3,35921253
$endgroup$
Normally I think of a morphism as belonging to two objects. e.g. for objects $a,b$ I think of a morphism $f$ between $a$ and $b$ as having the "type" $f:ato b$.
But I can't see in the definition of category that a morphism has to be connected specifically to two objects.
Can we have a category with distinct objects $a,b,c,d$ where $Hom(a,b)cap Hom(c,d)neq emptyset$?
example: we could have the morphisms be distances between objects where the objects are elements of a metric space. Then $Hom(a,b)= Hom(c,d)$ if $dist(a,b)=dist(c,d)$.
category-theory
category-theory
asked Jan 28 at 7:13
user56834user56834
3,35921253
asked Jan 28 at 7:13
user56834user56834
3,35921253
edited Jan 28 at 7:48
user56834
asked Jan 28 at 7:13
user56834user56834
3,35921253
asked Jan 28 at 7:13
user56834user56834
3,35921253
asked Jan 28 at 7:13
user56834user56834
3,35921253
3,35921253
$begingroup$
I like it when arrows in a category have well defined sources and targets....
$endgroup$
– Lord Shark the Unknown
Jan 28 at 7:24
add a comment |
$begingroup$
I like it when arrows in a category have well defined sources and targets....
$endgroup$
– Lord Shark the Unknown
Jan 28 at 7:24
$begingroup$
I like it when arrows in a category have well defined sources and targets....
$endgroup$
– Lord Shark the Unknown
Jan 28 at 7:24
$begingroup$
I like it when arrows in a category have well defined sources and targets....
$endgroup$
– Lord Shark the Unknown
Jan 28 at 7:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It depends on the definition of a category. If a category is defined as a $6$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),text{Mor}(mathcal{C}),text{dom}_{mathcal{C}},text{cod}_{mathcal{C}},circ_{mathcal{C}},text{id}_{mathcal{C}}),$$
where $text{Obj}(mathcal{C})$ and $text{Mor}(mathcal{C})$ are sets (classes) and $text{dom}_{mathcal{C}},text{cod}_{mathcal{C}}colontext{Mor}(mathcal{C})totext{Obj}(mathcal{C})$ are mappings, then all morphisms are disjoint and for objects $a,bintext{Obj}(mathcal{C})$ we have the following definition of the $hom$-set:
$$
text{hom}_{mathcal{C}}(a,b)={fintext{Mor}(mathcal{C})|text{dom}_{mathcal{C}}(f)=a, text{cod}_{mathcal{C}}(f)=b},
$$
so if $a,b,c,dintext{Obj}(mathcal{C})$ and $(a,b)ne(c,d)$, then
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)=varnothing.
$$
Now if your definition of a category is a $4$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})},(circ_{mathcal{C},a,b,c})_{a,b,cintext{Obj}(mathcal{C})},(text{id}_{mathcal{C},a})_{aintext{Obj}(mathcal{C})}),$$
where $text{Obj}(mathcal{C})$ is a set (class) and $(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})}$ is a family of sets, then it may happen that
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)nevarnothing.
$$
Moreover, all $hom$-sets may be equal. For example, if $X$ is a set, put $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={*}$ (fixed one-elemented set) for every $x_1,x_2in X$. Then for every $a,b,c,dintext{Obj}(mathcal{C})$ we have:
$$
hom_{mathcal{C}}(a,b)=hom_{mathcal{C}}(c,d).
$$
Your example with metric spaces also works (for a metric space $(X,d)$ we should take $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={d(x_1,x_2)}$; such categories are preorders where all elements are comparable; they are isomorphic to the categories from the previous example).
However, we may define a morphism as a triplet $(a,b,f)$, where $finhom_{mathcal{C}}(a,b)$, which turns the latter definition to the former (this is a simple construction, described in Mac Lane's "Categories for the Working Mathematician" at the bottom of the page 27). So it is just a matter of convention.
answered Jan 28 at 8:48
OskarOskar
3,2031819
$endgroup$
2
$begingroup$
I note for the benefit of the querent that even using the latter definition of category, once you start making any statements about composites you have to choose which composition operation you're applying, which gives a de facto unique source and target to any of the morphisms appearing in a particular statement. So even though you may have distinct hom-sets with a non-empty intersection, any category theoretic statement is going to work as if every morphism had a unique domain and codomain.
$endgroup$
– Malice Vidrine
Jan 28 at 9:38
add a comment |
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1 Answer
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$begingroup$
It depends on the definition of a category. If a category is defined as a $6$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),text{Mor}(mathcal{C}),text{dom}_{mathcal{C}},text{cod}_{mathcal{C}},circ_{mathcal{C}},text{id}_{mathcal{C}}),$$
where $text{Obj}(mathcal{C})$ and $text{Mor}(mathcal{C})$ are sets (classes) and $text{dom}_{mathcal{C}},text{cod}_{mathcal{C}}colontext{Mor}(mathcal{C})totext{Obj}(mathcal{C})$ are mappings, then all morphisms are disjoint and for objects $a,bintext{Obj}(mathcal{C})$ we have the following definition of the $hom$-set:
$$
text{hom}_{mathcal{C}}(a,b)={fintext{Mor}(mathcal{C})|text{dom}_{mathcal{C}}(f)=a, text{cod}_{mathcal{C}}(f)=b},
$$
so if $a,b,c,dintext{Obj}(mathcal{C})$ and $(a,b)ne(c,d)$, then
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)=varnothing.
$$
Now if your definition of a category is a $4$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})},(circ_{mathcal{C},a,b,c})_{a,b,cintext{Obj}(mathcal{C})},(text{id}_{mathcal{C},a})_{aintext{Obj}(mathcal{C})}),$$
where $text{Obj}(mathcal{C})$ is a set (class) and $(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})}$ is a family of sets, then it may happen that
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)nevarnothing.
$$
Moreover, all $hom$-sets may be equal. For example, if $X$ is a set, put $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={*}$ (fixed one-elemented set) for every $x_1,x_2in X$. Then for every $a,b,c,dintext{Obj}(mathcal{C})$ we have:
$$
hom_{mathcal{C}}(a,b)=hom_{mathcal{C}}(c,d).
$$
Your example with metric spaces also works (for a metric space $(X,d)$ we should take $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={d(x_1,x_2)}$; such categories are preorders where all elements are comparable; they are isomorphic to the categories from the previous example).
However, we may define a morphism as a triplet $(a,b,f)$, where $finhom_{mathcal{C}}(a,b)$, which turns the latter definition to the former (this is a simple construction, described in Mac Lane's "Categories for the Working Mathematician" at the bottom of the page 27). So it is just a matter of convention.
answered Jan 28 at 8:48
OskarOskar
3,2031819
$endgroup$
2
$begingroup$
I note for the benefit of the querent that even using the latter definition of category, once you start making any statements about composites you have to choose which composition operation you're applying, which gives a de facto unique source and target to any of the morphisms appearing in a particular statement. So even though you may have distinct hom-sets with a non-empty intersection, any category theoretic statement is going to work as if every morphism had a unique domain and codomain.
$endgroup$
– Malice Vidrine
Jan 28 at 9:38
add a comment |
$begingroup$
It depends on the definition of a category. If a category is defined as a $6$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),text{Mor}(mathcal{C}),text{dom}_{mathcal{C}},text{cod}_{mathcal{C}},circ_{mathcal{C}},text{id}_{mathcal{C}}),$$
where $text{Obj}(mathcal{C})$ and $text{Mor}(mathcal{C})$ are sets (classes) and $text{dom}_{mathcal{C}},text{cod}_{mathcal{C}}colontext{Mor}(mathcal{C})totext{Obj}(mathcal{C})$ are mappings, then all morphisms are disjoint and for objects $a,bintext{Obj}(mathcal{C})$ we have the following definition of the $hom$-set:
$$
text{hom}_{mathcal{C}}(a,b)={fintext{Mor}(mathcal{C})|text{dom}_{mathcal{C}}(f)=a, text{cod}_{mathcal{C}}(f)=b},
$$
so if $a,b,c,dintext{Obj}(mathcal{C})$ and $(a,b)ne(c,d)$, then
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)=varnothing.
$$
Now if your definition of a category is a $4$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})},(circ_{mathcal{C},a,b,c})_{a,b,cintext{Obj}(mathcal{C})},(text{id}_{mathcal{C},a})_{aintext{Obj}(mathcal{C})}),$$
where $text{Obj}(mathcal{C})$ is a set (class) and $(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})}$ is a family of sets, then it may happen that
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)nevarnothing.
$$
Moreover, all $hom$-sets may be equal. For example, if $X$ is a set, put $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={*}$ (fixed one-elemented set) for every $x_1,x_2in X$. Then for every $a,b,c,dintext{Obj}(mathcal{C})$ we have:
$$
hom_{mathcal{C}}(a,b)=hom_{mathcal{C}}(c,d).
$$
Your example with metric spaces also works (for a metric space $(X,d)$ we should take $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={d(x_1,x_2)}$; such categories are preorders where all elements are comparable; they are isomorphic to the categories from the previous example).
However, we may define a morphism as a triplet $(a,b,f)$, where $finhom_{mathcal{C}}(a,b)$, which turns the latter definition to the former (this is a simple construction, described in Mac Lane's "Categories for the Working Mathematician" at the bottom of the page 27). So it is just a matter of convention.
answered Jan 28 at 8:48
OskarOskar
3,2031819
$endgroup$
2
$begingroup$
I note for the benefit of the querent that even using the latter definition of category, once you start making any statements about composites you have to choose which composition operation you're applying, which gives a de facto unique source and target to any of the morphisms appearing in a particular statement. So even though you may have distinct hom-sets with a non-empty intersection, any category theoretic statement is going to work as if every morphism had a unique domain and codomain.
$endgroup$
– Malice Vidrine
Jan 28 at 9:38
add a comment |
$begingroup$
It depends on the definition of a category. If a category is defined as a $6$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),text{Mor}(mathcal{C}),text{dom}_{mathcal{C}},text{cod}_{mathcal{C}},circ_{mathcal{C}},text{id}_{mathcal{C}}),$$
where $text{Obj}(mathcal{C})$ and $text{Mor}(mathcal{C})$ are sets (classes) and $text{dom}_{mathcal{C}},text{cod}_{mathcal{C}}colontext{Mor}(mathcal{C})totext{Obj}(mathcal{C})$ are mappings, then all morphisms are disjoint and for objects $a,bintext{Obj}(mathcal{C})$ we have the following definition of the $hom$-set:
$$
text{hom}_{mathcal{C}}(a,b)={fintext{Mor}(mathcal{C})|text{dom}_{mathcal{C}}(f)=a, text{cod}_{mathcal{C}}(f)=b},
$$
so if $a,b,c,dintext{Obj}(mathcal{C})$ and $(a,b)ne(c,d)$, then
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)=varnothing.
$$
Now if your definition of a category is a $4$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})},(circ_{mathcal{C},a,b,c})_{a,b,cintext{Obj}(mathcal{C})},(text{id}_{mathcal{C},a})_{aintext{Obj}(mathcal{C})}),$$
where $text{Obj}(mathcal{C})$ is a set (class) and $(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})}$ is a family of sets, then it may happen that
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)nevarnothing.
$$
Moreover, all $hom$-sets may be equal. For example, if $X$ is a set, put $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={*}$ (fixed one-elemented set) for every $x_1,x_2in X$. Then for every $a,b,c,dintext{Obj}(mathcal{C})$ we have:
$$
hom_{mathcal{C}}(a,b)=hom_{mathcal{C}}(c,d).
$$
Your example with metric spaces also works (for a metric space $(X,d)$ we should take $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={d(x_1,x_2)}$; such categories are preorders where all elements are comparable; they are isomorphic to the categories from the previous example).
However, we may define a morphism as a triplet $(a,b,f)$, where $finhom_{mathcal{C}}(a,b)$, which turns the latter definition to the former (this is a simple construction, described in Mac Lane's "Categories for the Working Mathematician" at the bottom of the page 27). So it is just a matter of convention.
answered Jan 28 at 8:48
OskarOskar
3,2031819
$endgroup$
It depends on the definition of a category. If a category is defined as a $6$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),text{Mor}(mathcal{C}),text{dom}_{mathcal{C}},text{cod}_{mathcal{C}},circ_{mathcal{C}},text{id}_{mathcal{C}}),$$
where $text{Obj}(mathcal{C})$ and $text{Mor}(mathcal{C})$ are sets (classes) and $text{dom}_{mathcal{C}},text{cod}_{mathcal{C}}colontext{Mor}(mathcal{C})totext{Obj}(mathcal{C})$ are mappings, then all morphisms are disjoint and for objects $a,bintext{Obj}(mathcal{C})$ we have the following definition of the $hom$-set:
$$
text{hom}_{mathcal{C}}(a,b)={fintext{Mor}(mathcal{C})|text{dom}_{mathcal{C}}(f)=a, text{cod}_{mathcal{C}}(f)=b},
$$
so if $a,b,c,dintext{Obj}(mathcal{C})$ and $(a,b)ne(c,d)$, then
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)=varnothing.
$$
Now if your definition of a category is a $4$-tuple $$mathcal{C}=(text{Obj}(mathcal{C}),(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})},(circ_{mathcal{C},a,b,c})_{a,b,cintext{Obj}(mathcal{C})},(text{id}_{mathcal{C},a})_{aintext{Obj}(mathcal{C})}),$$
where $text{Obj}(mathcal{C})$ is a set (class) and $(hom_{mathcal{C}}(a,b))_{a,bintext{Obj}(mathcal{C})}$ is a family of sets, then it may happen that
$$
hom_{mathcal{C}}(a,b)caphom_{mathcal{C}}(c,d)nevarnothing.
$$
Moreover, all $hom$-sets may be equal. For example, if $X$ is a set, put $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={*}$ (fixed one-elemented set) for every $x_1,x_2in X$. Then for every $a,b,c,dintext{Obj}(mathcal{C})$ we have:
$$
hom_{mathcal{C}}(a,b)=hom_{mathcal{C}}(c,d).
$$
Your example with metric spaces also works (for a metric space $(X,d)$ we should take $text{Obj}(mathcal{C})=X$ and $hom_{mathcal{C}}(x_1,x_2)={d(x_1,x_2)}$; such categories are preorders where all elements are comparable; they are isomorphic to the categories from the previous example).
However, we may define a morphism as a triplet $(a,b,f)$, where $finhom_{mathcal{C}}(a,b)$, which turns the latter definition to the former (this is a simple construction, described in Mac Lane's "Categories for the Working Mathematician" at the bottom of the page 27). So it is just a matter of convention.
answered Jan 28 at 8:48
OskarOskar
3,2031819
edited Jan 28 at 8:54
answered Jan 28 at 8:48
OskarOskar
3,2031819
answered Jan 28 at 8:48
OskarOskar
3,2031819
answered Jan 28 at 8:48
OskarOskar
3,2031819
3,2031819
2
$begingroup$
I note for the benefit of the querent that even using the latter definition of category, once you start making any statements about composites you have to choose which composition operation you're applying, which gives a de facto unique source and target to any of the morphisms appearing in a particular statement. So even though you may have distinct hom-sets with a non-empty intersection, any category theoretic statement is going to work as if every morphism had a unique domain and codomain.
$endgroup$
– Malice Vidrine
Jan 28 at 9:38
add a comment |
2
$begingroup$
I note for the benefit of the querent that even using the latter definition of category, once you start making any statements about composites you have to choose which composition operation you're applying, which gives a de facto unique source and target to any of the morphisms appearing in a particular statement. So even though you may have distinct hom-sets with a non-empty intersection, any category theoretic statement is going to work as if every morphism had a unique domain and codomain.
$endgroup$
– Malice Vidrine
Jan 28 at 9:38
2
2
$begingroup$
I note for the benefit of the querent that even using the latter definition of category, once you start making any statements about composites you have to choose which composition operation you're applying, which gives a de facto unique source and target to any of the morphisms appearing in a particular statement. So even though you may have distinct hom-sets with a non-empty intersection, any category theoretic statement is going to work as if every morphism had a unique domain and codomain.
$endgroup$
– Malice Vidrine
Jan 28 at 9:38
$begingroup$
I note for the benefit of the querent that even using the latter definition of category, once you start making any statements about composites you have to choose which composition operation you're applying, which gives a de facto unique source and target to any of the morphisms appearing in a particular statement. So even though you may have distinct hom-sets with a non-empty intersection, any category theoretic statement is going to work as if every morphism had a unique domain and codomain.
$endgroup$
– Malice Vidrine
Jan 28 at 9:38
add a comment |
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$begingroup$
I like it when arrows in a category have well defined sources and targets....
$endgroup$
– Lord Shark the Unknown
Jan 28 at 7:24