Very basic question about pre-additive category
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I am trying to prove whether in a pre-additive category, $0_{Mor(y,z)}circ f=0_{Mor(x,z)}$ for objects $x,y,z$ and $fin Mor(x,y)$. Now, by bi-linearity of composition maps $$ 0_{Mor(y,z)}circ f+ 0_{Mor(y,z)}circ (-f)= 0_{Mor(y,z)} circ 0_{Mor(x,y)} $$
Putting $f=0_{Mor(x,y)}; ;$ we have $; ; 0_{Mor(y,z)} circ 0_{Mor(x,y)} =0_{Mor(x,z)}; ; $and we also get $; ; 0_{Mor(y,z)}circ f=-( 0_{Mor(y,z)}circ (-f))$. But I can't seem to progress any further.
Thanks in advance!
category-theory homological-algebra
asked Dec 31 '18 at 12:19
solgaleosolgaleo
9712
$endgroup$
add a comment |
$begingroup$
I am trying to prove whether in a pre-additive category, $0_{Mor(y,z)}circ f=0_{Mor(x,z)}$ for objects $x,y,z$ and $fin Mor(x,y)$. Now, by bi-linearity of composition maps $$ 0_{Mor(y,z)}circ f+ 0_{Mor(y,z)}circ (-f)= 0_{Mor(y,z)} circ 0_{Mor(x,y)} $$
Putting $f=0_{Mor(x,y)}; ;$ we have $; ; 0_{Mor(y,z)} circ 0_{Mor(x,y)} =0_{Mor(x,z)}; ; $and we also get $; ; 0_{Mor(y,z)}circ f=-( 0_{Mor(y,z)}circ (-f))$. But I can't seem to progress any further.
Thanks in advance!
category-theory homological-algebra
asked Dec 31 '18 at 12:19
solgaleosolgaleo
9712
$endgroup$
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
add a comment |
$begingroup$
I am trying to prove whether in a pre-additive category, $0_{Mor(y,z)}circ f=0_{Mor(x,z)}$ for objects $x,y,z$ and $fin Mor(x,y)$. Now, by bi-linearity of composition maps $$ 0_{Mor(y,z)}circ f+ 0_{Mor(y,z)}circ (-f)= 0_{Mor(y,z)} circ 0_{Mor(x,y)} $$
Putting $f=0_{Mor(x,y)}; ;$ we have $; ; 0_{Mor(y,z)} circ 0_{Mor(x,y)} =0_{Mor(x,z)}; ; $and we also get $; ; 0_{Mor(y,z)}circ f=-( 0_{Mor(y,z)}circ (-f))$. But I can't seem to progress any further.
Thanks in advance!
category-theory homological-algebra
asked Dec 31 '18 at 12:19
solgaleosolgaleo
9712
$endgroup$
I am trying to prove whether in a pre-additive category, $0_{Mor(y,z)}circ f=0_{Mor(x,z)}$ for objects $x,y,z$ and $fin Mor(x,y)$. Now, by bi-linearity of composition maps $$ 0_{Mor(y,z)}circ f+ 0_{Mor(y,z)}circ (-f)= 0_{Mor(y,z)} circ 0_{Mor(x,y)} $$
Putting $f=0_{Mor(x,y)}; ;$ we have $; ; 0_{Mor(y,z)} circ 0_{Mor(x,y)} =0_{Mor(x,z)}; ; $and we also get $; ; 0_{Mor(y,z)}circ f=-( 0_{Mor(y,z)}circ (-f))$. But I can't seem to progress any further.
Thanks in advance!
category-theory homological-algebra
category-theory homological-algebra
asked Dec 31 '18 at 12:19
solgaleosolgaleo
9712
asked Dec 31 '18 at 12:19
solgaleosolgaleo
9712
asked Dec 31 '18 at 12:19
solgaleosolgaleo
9712
asked Dec 31 '18 at 12:19
solgaleosolgaleo
9712
asked Dec 31 '18 at 12:19
solgaleosolgaleo
9712
9712
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
add a comment |
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
1
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
add a comment |
1 Answer
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$begingroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.4k11750
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$begingroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.4k11750
$endgroup$
add a comment |
$begingroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.4k11750
$endgroup$
add a comment |
$begingroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.4k11750
$endgroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.4k11750
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.4k11750
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.4k11750
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.4k11750
14.4k11750
add a comment |
add a comment |
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Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15