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Why is finiteness necessary in definition of connected category
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I am reading through Category Theory in Context by Emily Riehl. On page 33 she defines a connected category as one in which any pair of objects can be connected by a finite zigzag of morphisms (emphasis added).
Why do we need to specify a finite zigzag? Given an infinite family of morphisms from an object $x$ to an object $y,$ couldn't I just compose all of them to get a single morphism from $x$ to $y?$
My only guess is that "zig-zag" allows for morphisms pointing in opposite directions, e.g.
$$x rightarrow y leftarrow z$$
would constitute a zigzag of morphisms. In this case $x$ would be connected to $z$ (even if $textHom(x,z) = textHom(z,x) = emptyset$). On the other hand in the case of a countable family $y_i_i=1^infty$ of objects, the morphisms
$$x rightarrow y_1 leftarrow y_2 rightarrow cdots leftarrow z,$$
would not connect $x$ and $z$.
Is my interpretation correct or am I missing something?
category-theory
asked 4 hours ago
Aurel
406210
add a comment |Â
up vote
2
down vote
favorite
I am reading through Category Theory in Context by Emily Riehl. On page 33 she defines a connected category as one in which any pair of objects can be connected by a finite zigzag of morphisms (emphasis added).
Why do we need to specify a finite zigzag? Given an infinite family of morphisms from an object $x$ to an object $y,$ couldn't I just compose all of them to get a single morphism from $x$ to $y?$
My only guess is that "zig-zag" allows for morphisms pointing in opposite directions, e.g.
$$x rightarrow y leftarrow z$$
would constitute a zigzag of morphisms. In this case $x$ would be connected to $z$ (even if $textHom(x,z) = textHom(z,x) = emptyset$). On the other hand in the case of a countable family $y_i_i=1^infty$ of objects, the morphisms
$$x rightarrow y_1 leftarrow y_2 rightarrow cdots leftarrow z,$$
would not connect $x$ and $z$.
Is my interpretation correct or am I missing something?
category-theory
asked 4 hours ago
Aurel
406210
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading through Category Theory in Context by Emily Riehl. On page 33 she defines a connected category as one in which any pair of objects can be connected by a finite zigzag of morphisms (emphasis added).
Why do we need to specify a finite zigzag? Given an infinite family of morphisms from an object $x$ to an object $y,$ couldn't I just compose all of them to get a single morphism from $x$ to $y?$
My only guess is that "zig-zag" allows for morphisms pointing in opposite directions, e.g.
$$x rightarrow y leftarrow z$$
would constitute a zigzag of morphisms. In this case $x$ would be connected to $z$ (even if $textHom(x,z) = textHom(z,x) = emptyset$). On the other hand in the case of a countable family $y_i_i=1^infty$ of objects, the morphisms
$$x rightarrow y_1 leftarrow y_2 rightarrow cdots leftarrow z,$$
would not connect $x$ and $z$.
Is my interpretation correct or am I missing something?
category-theory
asked 4 hours ago
Aurel
406210
I am reading through Category Theory in Context by Emily Riehl. On page 33 she defines a connected category as one in which any pair of objects can be connected by a finite zigzag of morphisms (emphasis added).
Why do we need to specify a finite zigzag? Given an infinite family of morphisms from an object $x$ to an object $y,$ couldn't I just compose all of them to get a single morphism from $x$ to $y?$
My only guess is that "zig-zag" allows for morphisms pointing in opposite directions, e.g.
$$x rightarrow y leftarrow z$$
would constitute a zigzag of morphisms. In this case $x$ would be connected to $z$ (even if $textHom(x,z) = textHom(z,x) = emptyset$). On the other hand in the case of a countable family $y_i_i=1^infty$ of objects, the morphisms
$$x rightarrow y_1 leftarrow y_2 rightarrow cdots leftarrow z,$$
would not connect $x$ and $z$.
Is my interpretation correct or am I missing something?
category-theory
category-theory
asked 4 hours ago
Aurel
406210
asked 4 hours ago
Aurel
406210
asked 4 hours ago
Aurel
406210
asked 4 hours ago
Aurel
406210
asked 4 hours ago
Aurel
406210
406210
add a comment |Â
add a comment |Â
1 Answer
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Your guess is correct: the term "zig-zag" refers to a sequence of morphisms which could go in either direction. So, a zig-zag from $x$ to $y$ is a sequence of objects $z_0=x,z_1,z_2,dots,z_n=y$ together with either a morphism $z_ito z_i+1$ or a morphism $z_i+1to z_i$ for each $i$ from $0$ to $n-1$. The point of this definition is that "$x$ is connected to $z$" (i.e., "there exists a zig-zag from $x$ to $z$") should be the equivalence relation generated by "there exists a morphism from $x$ to $z$". If you imagine the category as a graph where the objects are vertices and the morphisms are edges, a zig-zag is a path from $x$ to $z$ (where we don't care about the directions of the arrows).
(This is a standard term in category theory, but its meaning certainly is not obvious, and I would consider it an error in the book that Riehl uses the term without first defining it.)
There is no such thing as an "infinite zig-zag" from $x$ to $y$ and it would not make sense to talk about such a thing. You can define an infinite zig-zag where your objects are indexed by $mathbbN$ (or $mathbbZ$), but in that case there would be no last (or first) object and so it would not make sense to say that the zig-zag is "from $x$ to $y$".
answered 4 hours ago
Eric Wofsey
168k12196312
Couldn't we well order the reals and index objects by that well-ordering? Then an object could have countably-infinite many predecessors?
– Aurel
4 hours ago
3
Well, sure, but that is highly unnatural, since there would be no relationship at all between $z_omega$ and $z_n$ for finite $n$, for instance. The point of this definition is that "$x$ is connected to $z$" should be the equivalence relation generated by "there exists a morphism from $x$ to $z$".
– Eric Wofsey
4 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Your guess is correct: the term "zig-zag" refers to a sequence of morphisms which could go in either direction. So, a zig-zag from $x$ to $y$ is a sequence of objects $z_0=x,z_1,z_2,dots,z_n=y$ together with either a morphism $z_ito z_i+1$ or a morphism $z_i+1to z_i$ for each $i$ from $0$ to $n-1$. The point of this definition is that "$x$ is connected to $z$" (i.e., "there exists a zig-zag from $x$ to $z$") should be the equivalence relation generated by "there exists a morphism from $x$ to $z$". If you imagine the category as a graph where the objects are vertices and the morphisms are edges, a zig-zag is a path from $x$ to $z$ (where we don't care about the directions of the arrows).
(This is a standard term in category theory, but its meaning certainly is not obvious, and I would consider it an error in the book that Riehl uses the term without first defining it.)
There is no such thing as an "infinite zig-zag" from $x$ to $y$ and it would not make sense to talk about such a thing. You can define an infinite zig-zag where your objects are indexed by $mathbbN$ (or $mathbbZ$), but in that case there would be no last (or first) object and so it would not make sense to say that the zig-zag is "from $x$ to $y$".
answered 4 hours ago
Eric Wofsey
168k12196312
Couldn't we well order the reals and index objects by that well-ordering? Then an object could have countably-infinite many predecessors?
– Aurel
4 hours ago
3
Well, sure, but that is highly unnatural, since there would be no relationship at all between $z_omega$ and $z_n$ for finite $n$, for instance. The point of this definition is that "$x$ is connected to $z$" should be the equivalence relation generated by "there exists a morphism from $x$ to $z$".
– Eric Wofsey
4 hours ago
add a comment |Â
up vote
5
down vote
accepted
Your guess is correct: the term "zig-zag" refers to a sequence of morphisms which could go in either direction. So, a zig-zag from $x$ to $y$ is a sequence of objects $z_0=x,z_1,z_2,dots,z_n=y$ together with either a morphism $z_ito z_i+1$ or a morphism $z_i+1to z_i$ for each $i$ from $0$ to $n-1$. The point of this definition is that "$x$ is connected to $z$" (i.e., "there exists a zig-zag from $x$ to $z$") should be the equivalence relation generated by "there exists a morphism from $x$ to $z$". If you imagine the category as a graph where the objects are vertices and the morphisms are edges, a zig-zag is a path from $x$ to $z$ (where we don't care about the directions of the arrows).
(This is a standard term in category theory, but its meaning certainly is not obvious, and I would consider it an error in the book that Riehl uses the term without first defining it.)
There is no such thing as an "infinite zig-zag" from $x$ to $y$ and it would not make sense to talk about such a thing. You can define an infinite zig-zag where your objects are indexed by $mathbbN$ (or $mathbbZ$), but in that case there would be no last (or first) object and so it would not make sense to say that the zig-zag is "from $x$ to $y$".
answered 4 hours ago
Eric Wofsey
168k12196312
Couldn't we well order the reals and index objects by that well-ordering? Then an object could have countably-infinite many predecessors?
– Aurel
4 hours ago
3
Well, sure, but that is highly unnatural, since there would be no relationship at all between $z_omega$ and $z_n$ for finite $n$, for instance. The point of this definition is that "$x$ is connected to $z$" should be the equivalence relation generated by "there exists a morphism from $x$ to $z$".
– Eric Wofsey
4 hours ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Your guess is correct: the term "zig-zag" refers to a sequence of morphisms which could go in either direction. So, a zig-zag from $x$ to $y$ is a sequence of objects $z_0=x,z_1,z_2,dots,z_n=y$ together with either a morphism $z_ito z_i+1$ or a morphism $z_i+1to z_i$ for each $i$ from $0$ to $n-1$. The point of this definition is that "$x$ is connected to $z$" (i.e., "there exists a zig-zag from $x$ to $z$") should be the equivalence relation generated by "there exists a morphism from $x$ to $z$". If you imagine the category as a graph where the objects are vertices and the morphisms are edges, a zig-zag is a path from $x$ to $z$ (where we don't care about the directions of the arrows).
(This is a standard term in category theory, but its meaning certainly is not obvious, and I would consider it an error in the book that Riehl uses the term without first defining it.)
There is no such thing as an "infinite zig-zag" from $x$ to $y$ and it would not make sense to talk about such a thing. You can define an infinite zig-zag where your objects are indexed by $mathbbN$ (or $mathbbZ$), but in that case there would be no last (or first) object and so it would not make sense to say that the zig-zag is "from $x$ to $y$".
answered 4 hours ago
Eric Wofsey
168k12196312
Your guess is correct: the term "zig-zag" refers to a sequence of morphisms which could go in either direction. So, a zig-zag from $x$ to $y$ is a sequence of objects $z_0=x,z_1,z_2,dots,z_n=y$ together with either a morphism $z_ito z_i+1$ or a morphism $z_i+1to z_i$ for each $i$ from $0$ to $n-1$. The point of this definition is that "$x$ is connected to $z$" (i.e., "there exists a zig-zag from $x$ to $z$") should be the equivalence relation generated by "there exists a morphism from $x$ to $z$". If you imagine the category as a graph where the objects are vertices and the morphisms are edges, a zig-zag is a path from $x$ to $z$ (where we don't care about the directions of the arrows).
(This is a standard term in category theory, but its meaning certainly is not obvious, and I would consider it an error in the book that Riehl uses the term without first defining it.)
There is no such thing as an "infinite zig-zag" from $x$ to $y$ and it would not make sense to talk about such a thing. You can define an infinite zig-zag where your objects are indexed by $mathbbN$ (or $mathbbZ$), but in that case there would be no last (or first) object and so it would not make sense to say that the zig-zag is "from $x$ to $y$".
answered 4 hours ago
Eric Wofsey
168k12196312
edited 4 hours ago
answered 4 hours ago
Eric Wofsey
168k12196312
answered 4 hours ago
Eric Wofsey
168k12196312
answered 4 hours ago
Eric Wofsey
168k12196312
168k12196312
Couldn't we well order the reals and index objects by that well-ordering? Then an object could have countably-infinite many predecessors?
– Aurel
4 hours ago
3
Well, sure, but that is highly unnatural, since there would be no relationship at all between $z_omega$ and $z_n$ for finite $n$, for instance. The point of this definition is that "$x$ is connected to $z$" should be the equivalence relation generated by "there exists a morphism from $x$ to $z$".
– Eric Wofsey
4 hours ago
add a comment |Â
Couldn't we well order the reals and index objects by that well-ordering? Then an object could have countably-infinite many predecessors?
– Aurel
4 hours ago
3
Well, sure, but that is highly unnatural, since there would be no relationship at all between $z_omega$ and $z_n$ for finite $n$, for instance. The point of this definition is that "$x$ is connected to $z$" should be the equivalence relation generated by "there exists a morphism from $x$ to $z$".
– Eric Wofsey
4 hours ago
Couldn't we well order the reals and index objects by that well-ordering? Then an object could have countably-infinite many predecessors?
– Aurel
4 hours ago
Couldn't we well order the reals and index objects by that well-ordering? Then an object could have countably-infinite many predecessors?
– Aurel
4 hours ago
3
3
Well, sure, but that is highly unnatural, since there would be no relationship at all between $z_omega$ and $z_n$ for finite $n$, for instance. The point of this definition is that "$x$ is connected to $z$" should be the equivalence relation generated by "there exists a morphism from $x$ to $z$".
– Eric Wofsey
4 hours ago
Well, sure, but that is highly unnatural, since there would be no relationship at all between $z_omega$ and $z_n$ for finite $n$, for instance. The point of this definition is that "$x$ is connected to $z$" should be the equivalence relation generated by "there exists a morphism from $x$ to $z$".
– Eric Wofsey
4 hours ago
add a comment |Â
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