What is the colon ':' called in category theory morphism definitions?












1












$begingroup$


Category theoretic morphism definition syntax, e.g.: $varphi : G to H$ uses the colon character rather than, for instance, the member-of sign e.g.: $varphi ∈ G to H$



What is the descriptive name of colon in this context? What is the semantic gain of this additional entity over member-of?






category-theory notation morphism







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    The function $varphi$ does not "equal" $G to H$ (whatever that means). There are probably lots of functions/morphisms $G to H$. This notation says that $varphi$ is one of them.
    $endgroup$
    – Randall
    Dec 17 '18 at 20:31












  • $begingroup$
    Roughly speaking, a colon can often be replaced with the phrase "such that". For instance, $varphi:Gto H$ reads "$varphi$ such that $G$ maps to $H$".
    $endgroup$
    – Shaun
    Dec 17 '18 at 20:35






  • 5




    $begingroup$
    @Shaun: People do occasionally use $:$ to mean 'such that', for example in set-builder notation (e.g. ${ x in mathbb{R} mid x ge 0 }$) or after an existential quantifier (e.g. $exists x in mathbb{R} : x ge 0$), but in the context of a function (or morphism in a category) that is neither what the colon means nor how it is read.
    $endgroup$
    – Clive Newstead
    Dec 17 '18 at 20:41








  • 3




    $begingroup$
    I've never heard anyone read this colon as "such that."
    $endgroup$
    – Randall
    Dec 17 '18 at 20:41
















1












$begingroup$


Category theoretic morphism definition syntax, e.g.: $varphi : G to H$ uses the colon character rather than, for instance, the member-of sign e.g.: $varphi ∈ G to H$



What is the descriptive name of colon in this context? What is the semantic gain of this additional entity over member-of?






category-theory notation morphism







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    The function $varphi$ does not "equal" $G to H$ (whatever that means). There are probably lots of functions/morphisms $G to H$. This notation says that $varphi$ is one of them.
    $endgroup$
    – Randall
    Dec 17 '18 at 20:31












  • $begingroup$
    Roughly speaking, a colon can often be replaced with the phrase "such that". For instance, $varphi:Gto H$ reads "$varphi$ such that $G$ maps to $H$".
    $endgroup$
    – Shaun
    Dec 17 '18 at 20:35






  • 5




    $begingroup$
    @Shaun: People do occasionally use $:$ to mean 'such that', for example in set-builder notation (e.g. ${ x in mathbb{R} mid x ge 0 }$) or after an existential quantifier (e.g. $exists x in mathbb{R} : x ge 0$), but in the context of a function (or morphism in a category) that is neither what the colon means nor how it is read.
    $endgroup$
    – Clive Newstead
    Dec 17 '18 at 20:41








  • 3




    $begingroup$
    I've never heard anyone read this colon as "such that."
    $endgroup$
    – Randall
    Dec 17 '18 at 20:41














1












1








1





$begingroup$


Category theoretic morphism definition syntax, e.g.: $varphi : G to H$ uses the colon character rather than, for instance, the member-of sign e.g.: $varphi ∈ G to H$



What is the descriptive name of colon in this context? What is the semantic gain of this additional entity over member-of?






category-theory notation morphism







share|cite|improve this question











$endgroup$




Category theoretic morphism definition syntax, e.g.: $varphi : G to H$ uses the colon character rather than, for instance, the member-of sign e.g.: $varphi ∈ G to H$



What is the descriptive name of colon in this context? What is the semantic gain of this additional entity over member-of?






category-theory notation morphism




category-theory notation morphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 15:16







James Bowery

















asked Dec 17 '18 at 20:30









James BoweryJames Bowery

1587




1587








  • 5




    $begingroup$
    The function $varphi$ does not "equal" $G to H$ (whatever that means). There are probably lots of functions/morphisms $G to H$. This notation says that $varphi$ is one of them.
    $endgroup$
    – Randall
    Dec 17 '18 at 20:31












  • $begingroup$
    Roughly speaking, a colon can often be replaced with the phrase "such that". For instance, $varphi:Gto H$ reads "$varphi$ such that $G$ maps to $H$".
    $endgroup$
    – Shaun
    Dec 17 '18 at 20:35






  • 5




    $begingroup$
    @Shaun: People do occasionally use $:$ to mean 'such that', for example in set-builder notation (e.g. ${ x in mathbb{R} mid x ge 0 }$) or after an existential quantifier (e.g. $exists x in mathbb{R} : x ge 0$), but in the context of a function (or morphism in a category) that is neither what the colon means nor how it is read.
    $endgroup$
    – Clive Newstead
    Dec 17 '18 at 20:41








  • 3




    $begingroup$
    I've never heard anyone read this colon as "such that."
    $endgroup$
    – Randall
    Dec 17 '18 at 20:41














  • 5




    $begingroup$
    The function $varphi$ does not "equal" $G to H$ (whatever that means). There are probably lots of functions/morphisms $G to H$. This notation says that $varphi$ is one of them.
    $endgroup$
    – Randall
    Dec 17 '18 at 20:31












  • $begingroup$
    Roughly speaking, a colon can often be replaced with the phrase "such that". For instance, $varphi:Gto H$ reads "$varphi$ such that $G$ maps to $H$".
    $endgroup$
    – Shaun
    Dec 17 '18 at 20:35






  • 5




    $begingroup$
    @Shaun: People do occasionally use $:$ to mean 'such that', for example in set-builder notation (e.g. ${ x in mathbb{R} mid x ge 0 }$) or after an existential quantifier (e.g. $exists x in mathbb{R} : x ge 0$), but in the context of a function (or morphism in a category) that is neither what the colon means nor how it is read.
    $endgroup$
    – Clive Newstead
    Dec 17 '18 at 20:41








  • 3




    $begingroup$
    I've never heard anyone read this colon as "such that."
    $endgroup$
    – Randall
    Dec 17 '18 at 20:41








5




5




$begingroup$
The function $varphi$ does not "equal" $G to H$ (whatever that means). There are probably lots of functions/morphisms $G to H$. This notation says that $varphi$ is one of them.
$endgroup$
– Randall
Dec 17 '18 at 20:31






$begingroup$
The function $varphi$ does not "equal" $G to H$ (whatever that means). There are probably lots of functions/morphisms $G to H$. This notation says that $varphi$ is one of them.
$endgroup$
– Randall
Dec 17 '18 at 20:31














$begingroup$
Roughly speaking, a colon can often be replaced with the phrase "such that". For instance, $varphi:Gto H$ reads "$varphi$ such that $G$ maps to $H$".
$endgroup$
– Shaun
Dec 17 '18 at 20:35




$begingroup$
Roughly speaking, a colon can often be replaced with the phrase "such that". For instance, $varphi:Gto H$ reads "$varphi$ such that $G$ maps to $H$".
$endgroup$
– Shaun
Dec 17 '18 at 20:35




5




5




$begingroup$
@Shaun: People do occasionally use $:$ to mean 'such that', for example in set-builder notation (e.g. ${ x in mathbb{R} mid x ge 0 }$) or after an existential quantifier (e.g. $exists x in mathbb{R} : x ge 0$), but in the context of a function (or morphism in a category) that is neither what the colon means nor how it is read.
$endgroup$
– Clive Newstead
Dec 17 '18 at 20:41






$begingroup$
@Shaun: People do occasionally use $:$ to mean 'such that', for example in set-builder notation (e.g. ${ x in mathbb{R} mid x ge 0 }$) or after an existential quantifier (e.g. $exists x in mathbb{R} : x ge 0$), but in the context of a function (or morphism in a category) that is neither what the colon means nor how it is read.
$endgroup$
– Clive Newstead
Dec 17 '18 at 20:41






3




3




$begingroup$
I've never heard anyone read this colon as "such that."
$endgroup$
– Randall
Dec 17 '18 at 20:41




$begingroup$
I've never heard anyone read this colon as "such that."
$endgroup$
– Randall
Dec 17 '18 at 20:41










2 Answers
2






active

oldest

votes


















5












$begingroup$

The colon is a typing symbol and does not represent equality.



The assertion $varphi : G to H$ asserts that $varphi$ is a morphism from the object $G$ to the object $H$. There may be many different morphisms from $G$ to $H$, so replacing $:$ by $=$ would not make much sense.



There are instances when a morphism is understood from context, when you might write an $=$ sign. For example, if you see something like:




...blah blah blah... and so we see trivially that $f = A to 1$.




What is probably meant is that $1$ is a terminal object in a category $mathcal{C}$ and $f$ is the unique morphism from $A$ to $1$ in $mathcal{C}$.



Related note: in the branch of type theory, the colon is used more widely; the notation $a : A$ represents the assertion that a term $a$ has type $A$. In type theory, the notation $A to B$ represents a function type, and then a function $f : A to B$ truly is a term $f$ of type $A to B$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    It's probably worth mentioning that with set-theoretic foundations, the notation $f:Ato B$ would typically be viewed as shorthand for $finmathsf{Hom}(A,B)$. The OP's question would have made a lot more sense if they'd contrasted to $in$ rather than $=$. There's probably a decent and not completely superficial discussion on why type theory uses $:$ over $in$ that could be had, but that's another question...
    $endgroup$
    – Derek Elkins
    Dec 17 '18 at 23:32










  • $begingroup$
    @DerekElkins: Good point. (For what it's worth, I have seen some presentations of type theory that use $in$ instead of $:$.)
    $endgroup$
    – Clive Newstead
    Dec 18 '18 at 2:12










  • $begingroup$
    I've corrected my question to use ∈ as the counter-example. Thanks for the feedback.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:16



















1












$begingroup$


The colon “:” generally abbreviates a pair of equalities.




A category is generally defined as consisting of things called ‘objects’ and
‘morphisms’ --as well as other things--
such that:
Each morphism $f$ has an associated “source” object and an associated “target” object.



To avoid being verbose, a syntactic construct is introduced:
$$
f : A → B qquad≡qquad mathsf{source},f ,=, A ;;land;; mathsf{target},f ,=, B
$$



Hope that helps :-)






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is extremely pedantic but technically simply knowing that $mathsf{source}(f)=A$ and $mathsf{target}(f)=B$ does not ensure that $f$ is a morphism of the category. You'd need an additional constraint like $finmathsf{Arr}(mathcal C)$.
    $endgroup$
    – Derek Elkins
    Dec 19 '18 at 2:13






  • 1




    $begingroup$
    I've corrected my question to use ∈ as the counter example. The syntactic sugar approach appeals to me but it leaves me a bit dissatisfied as it doesn't suggest a nice verbal translation e.g. "f morphs the category A to the category B" or something like that.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:21










  • $begingroup$
    $mathsf{source,target}$ are usual set-functions from the arrows of a category to its objects...
    $endgroup$
    – Musa Al-hassy
    Dec 19 '18 at 20:36











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

The colon is a typing symbol and does not represent equality.



The assertion $varphi : G to H$ asserts that $varphi$ is a morphism from the object $G$ to the object $H$. There may be many different morphisms from $G$ to $H$, so replacing $:$ by $=$ would not make much sense.



There are instances when a morphism is understood from context, when you might write an $=$ sign. For example, if you see something like:




...blah blah blah... and so we see trivially that $f = A to 1$.




What is probably meant is that $1$ is a terminal object in a category $mathcal{C}$ and $f$ is the unique morphism from $A$ to $1$ in $mathcal{C}$.



Related note: in the branch of type theory, the colon is used more widely; the notation $a : A$ represents the assertion that a term $a$ has type $A$. In type theory, the notation $A to B$ represents a function type, and then a function $f : A to B$ truly is a term $f$ of type $A to B$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    It's probably worth mentioning that with set-theoretic foundations, the notation $f:Ato B$ would typically be viewed as shorthand for $finmathsf{Hom}(A,B)$. The OP's question would have made a lot more sense if they'd contrasted to $in$ rather than $=$. There's probably a decent and not completely superficial discussion on why type theory uses $:$ over $in$ that could be had, but that's another question...
    $endgroup$
    – Derek Elkins
    Dec 17 '18 at 23:32










  • $begingroup$
    @DerekElkins: Good point. (For what it's worth, I have seen some presentations of type theory that use $in$ instead of $:$.)
    $endgroup$
    – Clive Newstead
    Dec 18 '18 at 2:12










  • $begingroup$
    I've corrected my question to use ∈ as the counter-example. Thanks for the feedback.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:16
















5












$begingroup$

The colon is a typing symbol and does not represent equality.



The assertion $varphi : G to H$ asserts that $varphi$ is a morphism from the object $G$ to the object $H$. There may be many different morphisms from $G$ to $H$, so replacing $:$ by $=$ would not make much sense.



There are instances when a morphism is understood from context, when you might write an $=$ sign. For example, if you see something like:




...blah blah blah... and so we see trivially that $f = A to 1$.




What is probably meant is that $1$ is a terminal object in a category $mathcal{C}$ and $f$ is the unique morphism from $A$ to $1$ in $mathcal{C}$.



Related note: in the branch of type theory, the colon is used more widely; the notation $a : A$ represents the assertion that a term $a$ has type $A$. In type theory, the notation $A to B$ represents a function type, and then a function $f : A to B$ truly is a term $f$ of type $A to B$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    It's probably worth mentioning that with set-theoretic foundations, the notation $f:Ato B$ would typically be viewed as shorthand for $finmathsf{Hom}(A,B)$. The OP's question would have made a lot more sense if they'd contrasted to $in$ rather than $=$. There's probably a decent and not completely superficial discussion on why type theory uses $:$ over $in$ that could be had, but that's another question...
    $endgroup$
    – Derek Elkins
    Dec 17 '18 at 23:32










  • $begingroup$
    @DerekElkins: Good point. (For what it's worth, I have seen some presentations of type theory that use $in$ instead of $:$.)
    $endgroup$
    – Clive Newstead
    Dec 18 '18 at 2:12










  • $begingroup$
    I've corrected my question to use ∈ as the counter-example. Thanks for the feedback.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:16














5












5








5





$begingroup$

The colon is a typing symbol and does not represent equality.



The assertion $varphi : G to H$ asserts that $varphi$ is a morphism from the object $G$ to the object $H$. There may be many different morphisms from $G$ to $H$, so replacing $:$ by $=$ would not make much sense.



There are instances when a morphism is understood from context, when you might write an $=$ sign. For example, if you see something like:




...blah blah blah... and so we see trivially that $f = A to 1$.




What is probably meant is that $1$ is a terminal object in a category $mathcal{C}$ and $f$ is the unique morphism from $A$ to $1$ in $mathcal{C}$.



Related note: in the branch of type theory, the colon is used more widely; the notation $a : A$ represents the assertion that a term $a$ has type $A$. In type theory, the notation $A to B$ represents a function type, and then a function $f : A to B$ truly is a term $f$ of type $A to B$.






share|cite|improve this answer









$endgroup$



The colon is a typing symbol and does not represent equality.



The assertion $varphi : G to H$ asserts that $varphi$ is a morphism from the object $G$ to the object $H$. There may be many different morphisms from $G$ to $H$, so replacing $:$ by $=$ would not make much sense.



There are instances when a morphism is understood from context, when you might write an $=$ sign. For example, if you see something like:




...blah blah blah... and so we see trivially that $f = A to 1$.




What is probably meant is that $1$ is a terminal object in a category $mathcal{C}$ and $f$ is the unique morphism from $A$ to $1$ in $mathcal{C}$.



Related note: in the branch of type theory, the colon is used more widely; the notation $a : A$ represents the assertion that a term $a$ has type $A$. In type theory, the notation $A to B$ represents a function type, and then a function $f : A to B$ truly is a term $f$ of type $A to B$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 20:37









Clive NewsteadClive Newstead

51.9k474136




51.9k474136








  • 2




    $begingroup$
    It's probably worth mentioning that with set-theoretic foundations, the notation $f:Ato B$ would typically be viewed as shorthand for $finmathsf{Hom}(A,B)$. The OP's question would have made a lot more sense if they'd contrasted to $in$ rather than $=$. There's probably a decent and not completely superficial discussion on why type theory uses $:$ over $in$ that could be had, but that's another question...
    $endgroup$
    – Derek Elkins
    Dec 17 '18 at 23:32










  • $begingroup$
    @DerekElkins: Good point. (For what it's worth, I have seen some presentations of type theory that use $in$ instead of $:$.)
    $endgroup$
    – Clive Newstead
    Dec 18 '18 at 2:12










  • $begingroup$
    I've corrected my question to use ∈ as the counter-example. Thanks for the feedback.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:16














  • 2




    $begingroup$
    It's probably worth mentioning that with set-theoretic foundations, the notation $f:Ato B$ would typically be viewed as shorthand for $finmathsf{Hom}(A,B)$. The OP's question would have made a lot more sense if they'd contrasted to $in$ rather than $=$. There's probably a decent and not completely superficial discussion on why type theory uses $:$ over $in$ that could be had, but that's another question...
    $endgroup$
    – Derek Elkins
    Dec 17 '18 at 23:32










  • $begingroup$
    @DerekElkins: Good point. (For what it's worth, I have seen some presentations of type theory that use $in$ instead of $:$.)
    $endgroup$
    – Clive Newstead
    Dec 18 '18 at 2:12










  • $begingroup$
    I've corrected my question to use ∈ as the counter-example. Thanks for the feedback.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:16








2




2




$begingroup$
It's probably worth mentioning that with set-theoretic foundations, the notation $f:Ato B$ would typically be viewed as shorthand for $finmathsf{Hom}(A,B)$. The OP's question would have made a lot more sense if they'd contrasted to $in$ rather than $=$. There's probably a decent and not completely superficial discussion on why type theory uses $:$ over $in$ that could be had, but that's another question...
$endgroup$
– Derek Elkins
Dec 17 '18 at 23:32




$begingroup$
It's probably worth mentioning that with set-theoretic foundations, the notation $f:Ato B$ would typically be viewed as shorthand for $finmathsf{Hom}(A,B)$. The OP's question would have made a lot more sense if they'd contrasted to $in$ rather than $=$. There's probably a decent and not completely superficial discussion on why type theory uses $:$ over $in$ that could be had, but that's another question...
$endgroup$
– Derek Elkins
Dec 17 '18 at 23:32












$begingroup$
@DerekElkins: Good point. (For what it's worth, I have seen some presentations of type theory that use $in$ instead of $:$.)
$endgroup$
– Clive Newstead
Dec 18 '18 at 2:12




$begingroup$
@DerekElkins: Good point. (For what it's worth, I have seen some presentations of type theory that use $in$ instead of $:$.)
$endgroup$
– Clive Newstead
Dec 18 '18 at 2:12












$begingroup$
I've corrected my question to use ∈ as the counter-example. Thanks for the feedback.
$endgroup$
– James Bowery
Dec 19 '18 at 15:16




$begingroup$
I've corrected my question to use ∈ as the counter-example. Thanks for the feedback.
$endgroup$
– James Bowery
Dec 19 '18 at 15:16











1












$begingroup$


The colon “:” generally abbreviates a pair of equalities.




A category is generally defined as consisting of things called ‘objects’ and
‘morphisms’ --as well as other things--
such that:
Each morphism $f$ has an associated “source” object and an associated “target” object.



To avoid being verbose, a syntactic construct is introduced:
$$
f : A → B qquad≡qquad mathsf{source},f ,=, A ;;land;; mathsf{target},f ,=, B
$$



Hope that helps :-)






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is extremely pedantic but technically simply knowing that $mathsf{source}(f)=A$ and $mathsf{target}(f)=B$ does not ensure that $f$ is a morphism of the category. You'd need an additional constraint like $finmathsf{Arr}(mathcal C)$.
    $endgroup$
    – Derek Elkins
    Dec 19 '18 at 2:13






  • 1




    $begingroup$
    I've corrected my question to use ∈ as the counter example. The syntactic sugar approach appeals to me but it leaves me a bit dissatisfied as it doesn't suggest a nice verbal translation e.g. "f morphs the category A to the category B" or something like that.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:21










  • $begingroup$
    $mathsf{source,target}$ are usual set-functions from the arrows of a category to its objects...
    $endgroup$
    – Musa Al-hassy
    Dec 19 '18 at 20:36
















1












$begingroup$


The colon “:” generally abbreviates a pair of equalities.




A category is generally defined as consisting of things called ‘objects’ and
‘morphisms’ --as well as other things--
such that:
Each morphism $f$ has an associated “source” object and an associated “target” object.



To avoid being verbose, a syntactic construct is introduced:
$$
f : A → B qquad≡qquad mathsf{source},f ,=, A ;;land;; mathsf{target},f ,=, B
$$



Hope that helps :-)






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is extremely pedantic but technically simply knowing that $mathsf{source}(f)=A$ and $mathsf{target}(f)=B$ does not ensure that $f$ is a morphism of the category. You'd need an additional constraint like $finmathsf{Arr}(mathcal C)$.
    $endgroup$
    – Derek Elkins
    Dec 19 '18 at 2:13






  • 1




    $begingroup$
    I've corrected my question to use ∈ as the counter example. The syntactic sugar approach appeals to me but it leaves me a bit dissatisfied as it doesn't suggest a nice verbal translation e.g. "f morphs the category A to the category B" or something like that.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:21










  • $begingroup$
    $mathsf{source,target}$ are usual set-functions from the arrows of a category to its objects...
    $endgroup$
    – Musa Al-hassy
    Dec 19 '18 at 20:36














1












1








1





$begingroup$


The colon “:” generally abbreviates a pair of equalities.




A category is generally defined as consisting of things called ‘objects’ and
‘morphisms’ --as well as other things--
such that:
Each morphism $f$ has an associated “source” object and an associated “target” object.



To avoid being verbose, a syntactic construct is introduced:
$$
f : A → B qquad≡qquad mathsf{source},f ,=, A ;;land;; mathsf{target},f ,=, B
$$



Hope that helps :-)






share|cite|improve this answer









$endgroup$




The colon “:” generally abbreviates a pair of equalities.




A category is generally defined as consisting of things called ‘objects’ and
‘morphisms’ --as well as other things--
such that:
Each morphism $f$ has an associated “source” object and an associated “target” object.



To avoid being verbose, a syntactic construct is introduced:
$$
f : A → B qquad≡qquad mathsf{source},f ,=, A ;;land;; mathsf{target},f ,=, B
$$



Hope that helps :-)







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answered Dec 18 '18 at 21:16









Musa Al-hassyMusa Al-hassy

1,3331711




1,3331711








  • 1




    $begingroup$
    This is extremely pedantic but technically simply knowing that $mathsf{source}(f)=A$ and $mathsf{target}(f)=B$ does not ensure that $f$ is a morphism of the category. You'd need an additional constraint like $finmathsf{Arr}(mathcal C)$.
    $endgroup$
    – Derek Elkins
    Dec 19 '18 at 2:13






  • 1




    $begingroup$
    I've corrected my question to use ∈ as the counter example. The syntactic sugar approach appeals to me but it leaves me a bit dissatisfied as it doesn't suggest a nice verbal translation e.g. "f morphs the category A to the category B" or something like that.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:21










  • $begingroup$
    $mathsf{source,target}$ are usual set-functions from the arrows of a category to its objects...
    $endgroup$
    – Musa Al-hassy
    Dec 19 '18 at 20:36














  • 1




    $begingroup$
    This is extremely pedantic but technically simply knowing that $mathsf{source}(f)=A$ and $mathsf{target}(f)=B$ does not ensure that $f$ is a morphism of the category. You'd need an additional constraint like $finmathsf{Arr}(mathcal C)$.
    $endgroup$
    – Derek Elkins
    Dec 19 '18 at 2:13






  • 1




    $begingroup$
    I've corrected my question to use ∈ as the counter example. The syntactic sugar approach appeals to me but it leaves me a bit dissatisfied as it doesn't suggest a nice verbal translation e.g. "f morphs the category A to the category B" or something like that.
    $endgroup$
    – James Bowery
    Dec 19 '18 at 15:21










  • $begingroup$
    $mathsf{source,target}$ are usual set-functions from the arrows of a category to its objects...
    $endgroup$
    – Musa Al-hassy
    Dec 19 '18 at 20:36








1




1




$begingroup$
This is extremely pedantic but technically simply knowing that $mathsf{source}(f)=A$ and $mathsf{target}(f)=B$ does not ensure that $f$ is a morphism of the category. You'd need an additional constraint like $finmathsf{Arr}(mathcal C)$.
$endgroup$
– Derek Elkins
Dec 19 '18 at 2:13




$begingroup$
This is extremely pedantic but technically simply knowing that $mathsf{source}(f)=A$ and $mathsf{target}(f)=B$ does not ensure that $f$ is a morphism of the category. You'd need an additional constraint like $finmathsf{Arr}(mathcal C)$.
$endgroup$
– Derek Elkins
Dec 19 '18 at 2:13




1




1




$begingroup$
I've corrected my question to use ∈ as the counter example. The syntactic sugar approach appeals to me but it leaves me a bit dissatisfied as it doesn't suggest a nice verbal translation e.g. "f morphs the category A to the category B" or something like that.
$endgroup$
– James Bowery
Dec 19 '18 at 15:21




$begingroup$
I've corrected my question to use ∈ as the counter example. The syntactic sugar approach appeals to me but it leaves me a bit dissatisfied as it doesn't suggest a nice verbal translation e.g. "f morphs the category A to the category B" or something like that.
$endgroup$
– James Bowery
Dec 19 '18 at 15:21












$begingroup$
$mathsf{source,target}$ are usual set-functions from the arrows of a category to its objects...
$endgroup$
– Musa Al-hassy
Dec 19 '18 at 20:36




$begingroup$
$mathsf{source,target}$ are usual set-functions from the arrows of a category to its objects...
$endgroup$
– Musa Al-hassy
Dec 19 '18 at 20:36


















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