# Help on Category Theory

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• September 13th 2009, 05:25 PM
redsoxfan325
Help on Category Theory
I'm taking an advanced abstract algebra course this year, and the professor decided to give us some homework on category theory without actually teaching it to us. The next time we meet (Wednesday) is when the homework is due.

I think I understand what a morphism is: A set of all maps from one object to another (in a given category). I understand the definition of initial and final objects, but I can't think of any examples of them OR how to be able to find them given a category. For example, one question on the homework is "Indicate the initial and final objects of the category $Vect(K)$ of finite-dimensional vector spaces over the field $K$."

Two other related concepts I don't understand are that of a direct sum and canonical embeddings. We need to show that in the category $Vect(K)$, direct sums exists for all pairs of objects.

• September 13th 2009, 10:40 PM
Jose27
Quote:

Originally Posted by redsoxfan325
I'm taking an advanced abstract algebra course this year, and the professor decided to give us some homework on category theory without actually teaching it to us. The next time we meet (Wednesday) is when the homework is due.

I think I understand what a morphism is: A set of all maps from one object to another (in a given category). I understand the definition of initial and final objects, but I can't think of any examples of them OR how to be able to find them given a category. For example, one question on the homework is "Indicate the initial and final objects of the category $Vect(K)$ of finite-dimensional vector spaces over the field $K$."

Two other related concepts I don't understand are that of a direct sum and canonical embeddings. We need to show that in the category $Vect(K)$, direct sums exists for all pairs of objects.

A morphism is an arrow from one object to another, not the whole set (assuming your category is locally small).

Remember that an initial object is an object $A$ in your category $\mathcal{C}$ such that for every object $B$ in $\mathcal{C}$ there is a unique morphism $f:A \longrightarrow B$. If you're in $\mathcal{VECT(K)}$ then $\{ 0 \}$ clearly satisfies the definition, since a morphism in $\mathcal{VECT(K)}$ is a linear transformation. Similarly you can show that $\{ 0 \}$ is the terminal object.

For the second part I haven't reached those topics so I really can't say.
• September 14th 2009, 03:26 AM
NonCommAlg
the product of two objects $A,B$ in a category $\mathcal{C}$ is an object $C$ and morphisms $p_1: C \longrightarrow A, \ p_2: C \longrightarrow B$ such that for every object $D$ and morphisms $f: D \longrightarrow A$

and $g: D \longrightarrow B,$ there exists a unique morphism $h: D \longrightarrow C$ such that our diagram commutes, i.e. $p_1h=f, \ p_2h=g.$

the reason that i showed the morphisms by $p_1,p_2$ is that they really behave like "projections". anyway, it's not hard to prove that the direct sum (or product) is unique if it

exists. for the category $\mathcal{VECT}(K)$ define $C=A \oplus B,$ where $\oplus$ is the ordinary direct product of vector spaces. also define $p_1(a,b)=a, \ p_2(a,b)=b,$ for all $a \in A, \ b \in B.$

now for a (finite dimensional) $K$-vector space $D$ and linear transformations $f: D \longrightarrow A, \ g: D \longrightarrow B$ define $h: D \longrightarrow A \oplus B$ by $h(d)=(f(d),g(d)).$ clearly $h$ is a linear

transformation because $f,g$ are and $p_1h=f, \ p_2h=g.$ if $u: D \longrightarrow A \oplus B$ is another such linear transformation, then assuming that $u(d)=(\alpha(d), \beta(d)),$ we will have

$f(d)=p_1u(d)=\alpha(d), \ g(d)=p_2u(d)=\beta(d).$ thus $u = h.$ so $h$ is unique and we're done.

Edit: Ok, i just noticed that your question was about "direct sum", which i think you mean "coproduct". this is just the "dual" of the product. so you just reverse the direction

of all arrows to get the definition of coproduct. for the category K-vector spaces (more generally R-modules) product and coproduct of a finitely many of objects are the same.
• September 14th 2009, 06:13 AM
aliceinwonderland
Quote:

Originally Posted by NonCommAlg
the product of two objects $A,B$ in a category $\mathcal{C}$ is an object $C$ and morphisms $p_1: C \longrightarrow A, \ p_2: C \longrightarrow B$ such that for every object $D$ and morphisms $f: D \longrightarrow A$

and $g: D \longrightarrow B,$ there exists a unique morphism $h: D \longrightarrow C$ such that our diagram commutes, i.e. $p_1h=f, \ p_2h=g.$

the reason that i showed the morphisms by $p_1,p_2$ is that they really behave like "projections". anyway, it's not hard to prove that the direct sum (or product) is unique if it

exists. for the category $\mathcal{VECT}(K)$ define $C=A \oplus B,$ where $\oplus$ is the ordinary direct product of vector spaces. also define $p_1(a,b)=a, \ p_2(a,b)=b,$ for all $a \in A, \ b \in B.$

now for a (finite dimensional) $K$-vector space $D$ and linear transformations $f: D \longrightarrow A, \ g: D \longrightarrow B$ define $h: D \longrightarrow A \oplus B$ by $h(d)=(f(d),g(d)).$ clearly $h$ is a linear

transformation because $f,g$ are and $p_1h=f, \ p_2h=g.$ if $u: D \longrightarrow A \oplus B$ is another such linear transformation, then assuming that $u(d)=(\alpha(d), \beta(d)),$ we will have

$f(d)=p_1u(d)=\alpha(d), \ g(d)=p_2u(d)=\beta(d).$ thus $u = h.$ so $h$ is unique and we're done.

Edit: Ok, i just noticed that your question was about "direct sum", which i think you mean "coproduct". this is just the "dual" of the product. so you just reverse the direction

of all arrows to get the definition of coproduct. for the category K-vector spaces (more generally R-modules) product and coproduct of a finitely many of objects are the same.

In wiki,
".. in the category of abelian groups (and equally for vector spaces), the coproduct, called the direct sum, consists of the elements of the direct product which have only "finitely" many nonzero terms (this therefore coincides exactly with the direct product, in the case of finitely many factors). "

If we drop the "finitely many" condition for the direct product of abelian groups, why it is not the coproduct of the category of abelian groups?
• September 14th 2009, 09:30 AM
NonCommAlg
Quote:

Originally Posted by aliceinwonderland

If we drop the "finitely many" condition for the direct product of abelian groups, why it is not the coproduct of the category of abelian groups?

before i answer your question let me give the definition of product and coproduct for the most general case, i.e. for a "family" $\{A_j \}_{j \in I}$ of objects in a category $\mathcal{C}$:

1) the product is an objcet $B$ and morphims $p_j: B \longrightarrow A_j, \ j \in I,$ such that for every object $D$ and morphisms $f_j : D \longrightarrow A_j, \ j \in I,$ there exists a unique morphism

$g: D \longrightarrow B$ such that $p_jg=f_j, \ j \in I.$

2) the coproduct is an object $C$ and morphisms $\iota_j: A_j : \longrightarrow C, \ j \in I,$ such that for every object $D$ and morphisms $f_j : A_j \longrightarrow D, \ j\in I,$ there exists a unique morphism

$g: C \longrightarrow D$ such that $g\iota_j=f_j, \ j \in I.$

from now on i'll assume that $\mathcal{C}$ is the category of all vector spaces over a field $K$ [or the category of abelian groups or in general the category of $R$ modules, where $R$ is a

(commutative) ring]. then, exactly as i showed in my previous post, $\prod_{j \in I} A_j$ will be the product of $A_j, \ j \in I.$ clearly if $\mathcal{C}$ is the category of finite dimensional vector spaces over

a field $K,$ then the index set $I$ has to be finite because otherwise the vector space $\prod_{j \in I} A_j$ will not be in $\mathcal{C}$ anymore.

for the coproduct let $C=\oplus_{j \in I} A_j$ and define $\iota_j : A_j \longrightarrow C, \ j \in I,$ to be the natural injection. let $p_j: C \longrightarrow A_j, \ j \in I,$ be the natural projection. now for any $D \in \mathcal{C},$ and

morphisms $f_j : A_j \longrightarrow D$ define $g: C \longrightarrow D$ by $g(a) = \sum_{i \in I} f_ip_i(a).$ this definition is well-defined only if only we have finitely many term in the sum, which explains why we've

considered $\oplus$ instead of $\prod.$ (this also answers aliceinwonderland's question!) now for any $j \in I$ we have $g \iota_j(a_j)=\sum_{i \in I} f_ip_i(\iota_j(a_j))=f_jp_j(\iota_j(a_j))=f_j(a_j) .$ thus $g \iota_j = f_j$

for all $j \in I.$ so, in order to prove that $C=\oplus_{j \in I} A_j$ is the coproduct of $A_j, \ j \in I,$ we only need to prove the uniqueness of $g,$ which is easy.

again, note that if $\mathcal{C}$ is the category of finite dimensional vector spaces over some field $K,$ then the index set $I$ will have to be finite.
• September 14th 2009, 10:13 AM
aliceinwonderland
Quote:

Originally Posted by NonCommAlg
before i answer your question let me give the definition of product and coproduct for the most general case, i.e. for a "family" $\{A_j \}_{j \in I}$ of objects in a category $\mathcal{C}$:

1) the product is an objcet $B$ and morphims $p_j: B \longrightarrow A_j, \ j \in I,$ such that for every object $D$ and morphisms $f_j : D \longrightarrow A_j, \ j \in I,$ there exists a unique morphism

$g: D \longrightarrow B$ such that $p_jg=f_j, \ j \in I.$

2) the coproduct is an object $C$ and morphisms $\iota_j: A_j : \longrightarrow C, \ j \in I,$ such that for every object $D$ and morphisms $f_j : A_j \longrightarrow D, \ j\in I,$ there exists a unique morphism

$g: C \longrightarrow D$ such that $g\iota_j=f_j, \ j \in I.$

from now on i'll assume that $\mathcal{C}$ is the category of all vector spaces over a field $K$ [or the category of abelian groups or in general the category of $R$ modules, where $R$ is a

(commutative) ring]. then, exactly as i showed in my previous post, $\prod_{j \in I} A_j$ will be the product of $A_j, \ j \in I.$ clearly if $\mathcal{C}$ is the category of finite dimensional vector spaces over

a field $K,$ then the index set $I$ has to be finite because otherwise the vector space $\prod_{j \in I} A_j$ will not be in $\mathcal{C}$ anymore.

for the coproduct let $C=\oplus_{j \in I} A_j$ and define $\iota_j : A_j \longrightarrow C, \ j \in I,$ to be the natural injection. let $p_j: C \longrightarrow A_j, \ j \in I,$ be the natural projection. now for any $D \in \mathcal{C},$ and

morphisms $f_j : A_j \longrightarrow D$ define $g: C \longrightarrow D$ by $g(a) = \sum_{i \in I} f_ip_i(a).$ this definition is well-defined only if only we have finitely many term in the sum, which explains why we've

considered $\oplus$ instead of $\prod.$ (this also answers aliceinwonderland's question!) now for any $j \in I$ we have $g \iota_j(a_j)=\sum_{i \in I} f_ip_i(\iota_j(a_j))=f_jp_j(\iota_j(a_j))=f_j(a_j) .$ thus $g \iota_j = f_j$

for all $j \in I.$ so, in order to prove that $C=\oplus_{j \in I} A_j$ is the coproduct of $A_j, \ j \in I,$ we only need to prove the uniqueness of $g,$ which is easy.

again, note that if $\mathcal{C}$ is the category of finite dimensional vector spaces over some field $K,$ then the index set $I$ will have to be finite.

I understand that if the category C is the category of finitely generated abelian groups, then both direct product and direct sum can be a coproduct in C.

If Ab is the category of abelian groups in a general sense, why the direct product is not a coproduct in Ab? Any example?

Thanks.
• September 14th 2009, 06:34 PM
NonCommAlg
Quote:

Originally Posted by aliceinwonderland

I understand that if the category C is the category of finitely generated abelian groups, then both direct product and direct sum can be a coproduct in C.

yes, because for the category of f.g. abelian groups, the product (coproduct) is guaranteed to exist for a finite set of the objects of that category and obviously $\bigoplus_{j=1}^n A_j \cong \prod_{j=1}^n A_j.$

Quote:

If Ab is the category of abelian groups in a general sense, why the direct product is not a coproduct in Ab? Any example?

Thanks.
if the coproduct (or product) of a family of objects in a category exists, it'll have to be unique. for $\text{Ab},$ as we showed, the direct sum is the coproduct. so your question can be rephrased

in this way: can we find an example of a family of abelian groups $A_j, \ j \in I,$ such that $\bigoplus_{j \in I} A_j \ncong \prod_{j \in I} A_j$ ? the answer is yes, there are millions of such examples. here's an obvious one:

let $A_j = \mathbb{Z}/j \mathbb{Z}, \ j \in \mathbb{N}.$ then every element of $\bigoplus_{j=1}^{\infty} A_j$ has finite order but this is clearly not true in $\prod_{j=1}^{\infty} A_j.$
• September 15th 2009, 12:23 PM
redsoxfan325
Quote:

Originally Posted by NonCommAlg
the product of two objects $A,B$ in a category $\mathcal{C}$ is an object $C$ and morphisms $p_1: C \longrightarrow A, \ p_2: C \longrightarrow B$ such that for every object $D$ and morphisms $f: D \longrightarrow A$

and $g: D \longrightarrow B,$ there exists a unique morphism $h: D \longrightarrow C$ such that our diagram commutes, i.e. $p_1h=f, \ p_2h=g.$

the reason that i showed the morphisms by $p_1,p_2$ is that they really behave like "projections". anyway, it's not hard to prove that the direct sum (or product) is unique if it

exists. for the category $\mathcal{VECT}(K)$ define $C=A \oplus B,$ where $\oplus$ is the ordinary direct product of vector spaces. also define $p_1(a,b)=a, \ p_2(a,b)=b,$ for all $a \in A, \ b \in B.$

[color="blue"]now for a (finite dimensional) $K$-vector space $D$ and linear transformations $f: D \longrightarrow A, \ g: D \longrightarrow B$ define $h: D \longrightarrow A \oplus B$ by $h(d)=(f(d),g(d)).$ clearly $h$ is a linear

transformation because $f,g$ are and $p_1h=f, \ p_2h=g.$ if $u: D \longrightarrow A \oplus B$ is another such linear transformation, then assuming that $u(d)=(\alpha(d), \beta(d)),$ we will have

$f(d)=p_1u(d)=\alpha(d), \ g(d)=p_2u(d)=\beta(d).$ thus $u = h.$ so $h$ is unique and we're done.

Edit: Ok, i just noticed that your question was about "direct sum", which i think you mean "coproduct". this is just the "dual" of the product. so you just reverse the direction

of all arrows to get the definition of coproduct. for the category K-vector spaces (more generally R-modules) product and coproduct of a finitely many of objects are the same.

OK, I've spent the last day trying to wrap my head around this. You say that for direct sum I should just reverse the arrows, but the functions don't really make sense any more, do they? For instance, $p_1:C\longrightarrow A$. You say define $p_1(a,b)=a$, which is fine, but if I want $p_1: A\longrightarrow C$, what is $p_1$ now?

If I have two vector spaces $U$ and $V$ what would morphisms $i:U\longrightarrow (U\oplus V)$ and $j:V\longrightarrow (U\oplus V)$ be? Or what about $\phi: (U\oplus V)\longrightarrow W$, where $W$ is another vector space? I am really struggling with this.
• September 15th 2009, 12:28 PM
redsoxfan325
I've attached a picture of the problem below.
• September 15th 2009, 06:55 PM
NonCommAlg
Quote:

Originally Posted by redsoxfan325
OK, I've spent the last day trying to wrap my head around this. You say that for direct sum I should just reverse the arrows, but the functions don't really make sense any more, do they? For instance, $p_1:C\longrightarrow A$. You say define $p_1(a,b)=a$, which is fine, but if I want $p_1: A\longrightarrow C$, what is $p_1$ now?

you're right that just reversing the direction of the arrows is not enough. as i explained in # 5 in this thread, the morphisms this time will be the natural injections $\iota_1: A_1 \longrightarrow C$ and $\iota_2 : A_2 \longrightarrow C,$

where $C=A_1 \oplus A_2$ and $\iota_1, \ \iota_2$ are defined by $\iota_1(a_1)=(a_1,0), \ \iota_2(a_2)=(0,a_2),$ for all $a_1 \in A_1, \ a_2 \in A_2.$ now if $A_1,A_2$ are, say (finite dimensional) vector spaces over some field $K,$ then $C$ will

also be a (finite dimensional) vector space over $K,$ i.e. $C$ is an object in our category. also it's clear that $\iota_1, \iota_2$ are $K$-linear transformations, i.e. they are morphisms in our category. to prove that

$C$ is a coproduct of $A_1,A_2,$ we need to prove that for any (finite dimensional) $K$-vector space $D$ and any linear transformations $f_j : A_j \longrightarrow D, \ j=1,2,$ there exists a unique linear transformation
$g: C \longrightarrow D$ with this property that $g \iota_j = f_j, \ j=1,2.$ here's how to prove it:

existence of $g$: define $g: C \longrightarrow D$ by $g(a)=f_1p_1(a) + f_2p_2(a),$ where $p_j : C \longrightarrow A_j, \ j=1,2,$ are the natural projections. it's clear that $g$ is a linear transformation because $f_j,p_j$ are. so we

now need to show that $g \iota_j = f_j, \ j=1,2.$ so suppose $a_1 \in A_1.$ then $g \iota_1 (a_1)=g(a_1,0)=f_1p_1(a_1,0) + f_2p_2(a_1,0)=f_1(a_1) + f_2(0)=f_1(a_1).$ similarly we see that $g \iota_2 = f_2.$

uniqueness of $g$: suppose $h: C \longrightarrow D$ is any morphism with this property that $h \iota_j = f_j, \ j=1,2.$ we need to show that $h=g$: let $a=(a_1,a_2) \in C.$ then:

$h(a)=h(a_1,0) + h(0,a_2)=h \iota_1(a_1) + h \iota_2(a_2)=f_1(a_1)+f_2(a_2).$ but we have $a_1=p_1(a), \ a_2 = p_2(a)$ and thus $h(a)=f_1p_1(a) + f_2p_2(a) = g(a). \ \Box$

Quote:

Originally Posted by redsoxfan325

I've attached a picture of the problem below.

this definition of coproduct (direct sum) is exactly the one that i gave you but in a more categorical language. the definition that i gave you is more standard.
• September 15th 2009, 08:30 PM
redsoxfan325
Quote:

Originally Posted by NonCommAlg
you're right that just reversing the direction of the arrows is not enough. as i explained in # 5 in this thread, the morphisms this time will be the natural injections $\iota_1: A_1 \longrightarrow C$ and $\iota_2 : A_2 \longrightarrow C,$

where $C=A_1 \oplus A_2$ and $\iota_1, \ \iota_2$ are defined by $\iota_1(a_1)=(a_1,0), \ \iota_2(a_2)=(0,a_2),$ for all $a_1 \in A_1, \ a_2 \in A_2.$ now if $A_1,A_2$ are, say (finite dimensional) vector spaces over some field $K,$ then $C$ will

also be a (finite dimensional) vector space over $K,$ i.e. $C$ is an object in our category. also it's clear that $\iota_1, \iota_2$ are $K$-linear transformations, i.e. they are morphisms in our category. to prove that

$C$ is a coproduct of $A_1,A_2,$ we need to prove that for any (finite dimensional) $K$-vector space $D$ and any linear transformations $f_j : A_j \longrightarrow D, \ j=1,2,$ there exists a unique linear transformation
$g: C \longrightarrow D$ with this property that $g \iota_j = f_j, \ j=1,2.$ here's how to prove it:

existence of $g$: define $g: C \longrightarrow D$ by $g(a)=f_1p_1(a) + f_2p_2(a),$ where $p_j : C \longrightarrow A_j, \ j=1,2,$ are the natural projections. it's clear that $g$ is a linear transformation because $f_j,p_j$ are. so we

now need to show that $g \iota_j = f_j, \ j=1,2.$ so suppose $a_1 \in A_1.$ then $g \iota_1 (a_1)=g(a_1,0)=f_1p_1(a_1,0) + f_2p_2(a_1,0)=f_1(a_1) + f_2(0)=f_1(a_1).$ similarly we see that $g \iota_2 = f_2.$

uniqueness of $g$: suppose $h: C \longrightarrow D$ is any morphism with this property that $h \iota_j = f_j, \ j=1,2.$ we need to show that $h=g$: let $a=(a_1,a_2) \in C.$ then:

$h(a)=h(a_1,0) + h(0,a_2)=h \iota_1(a_1) + h \iota_2(a_2)=f_1(a_1)+f_2(a_2).$ but we have $a_1=p_1(a), \ a_2 = p_2(a)$ and thus $h(a)=f_1p_1(a) + f_2p_2(a) = g(a). \ \Box$

this definition of coproduct (direct sum) is exactly the one that i gave you but in a more categorical language. the definition that i gave you is more standard.

Thank you very much.
• November 12th 2009, 10:49 PM
ctnovice
Basic category theory help
Dear 'aliceinwonderland',

I'm sorry to bother you. I'm really desperate for help with some category theory questions. Would you mind if I emailed you with details?

Thanks.

Joe
• November 13th 2009, 10:19 AM
aliceinwonderland
Quote:

Originally Posted by ctnovice
Dear 'aliceinwonderland',

I'm sorry to bother you. I'm really desperate for help with some category theory questions. Would you mind if I emailed you with details?

Thanks.

Joe

Well, I am not an expert of category theory.
Why don't you post your question here? Some people (including me) might be able to help you. If you question includes lots of arrows and diagrams, just attach a file to your post.
• November 15th 2009, 10:20 AM
ctnovice
ok sure, thanks for the advice.
• November 16th 2009, 08:58 PM
aliceinwonderland
Quote:

Originally Posted by ctnovice
Could anyone please help with a basic category question I'm struggling with. I'm quite new to this world.

Please see the attached document for the problem - "joe_question_a.doc"

Joe.

Let J be an index category and Sets for the category of sets; define a functor F:J-->Sets that maps every arrow of J to an inclusion. We see that F is a nested sequence of sets for your question such that $(F_1 = X_1) \subset (F_2 = X_2) \subset$....
Colimit of (a) is the union U of all sets X_n with cocone given by inclusion maps $\phi_n : (X_n) \rightarrow U$, where $\phi_n^{-1}=e_n^{-1} \cdot e_{n+1}^{-1} \cdots$

For (b), the infinite product of X_n, $\prod_j{ X_j}$, with cone given by projection maps $q_j : \prod_j{X_j} \rightarrow X_j$, where $q_n^{-1}=p_n^{-1} \cdot p_{n+1}^{-1}, \cdots$

To show that a universal property for (a), given cocone $(U, \phi)$ and another cocone $(V, \psi)$, there exists a unique morphism $f:V \rightarrow U$ such that $f\psi_j = \phi_j$ for all j in J.

To show that a universal property for (b), given cone $(\prod_j{ X_j}, q)$ and another cone $(C, y)$, there exists a unique morphism $f:\prod_j{ X_j} \rightarrow C$ such that $y_jf = q_j$ for all j in J.

Note: I don't guarantee that the above solution of mine is error-free. Compare your own solution and let me know if you find any error.
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