Exercise $1.1.4$, Weibel
$begingroup$
I'm new to homological algebra, and I'm trying to learn it using Weibel's An introduction to homological algebra. I'm currently having trouble with exercise 1.1.4:
Show that ${operatorname{Hom}_R(A, C_n)}$ forms a chain complex of abelian groups for every $R$-module $A$ and every $R$-module chain complex $C$. Taking $A=Z_n$ show that if $H_n(operatorname{Hom}_R(Z_n,C))=0$, then $H_n(C)=0$. Is the converse true?
I think the first part is OK, using that if the chain complex looks like:
$$C:~ldotsrightarrow C_noverset{d_n}{rightarrow}C_{n-1}rightarrowldots$$
Then we get a new chain complex:
$$C^*:~ldotsrightarrowoperatorname{Hom}(A,C_n)overset{d^*_n}{rightarrow}operatorname{Hom}(A,C_{n-1})rightarrowldots$$
Where $d^*(f_n)=d_ncirc f_n$ for a $f_ninoperatorname{Hom}(A,C_n)$. This is a chain complex since $d^*_{n-1}circ d^*_n(f_n)=(d_{n-1}circ d_n)(f_n)=0$.
I'm not so sure about the second part. Assuming $Z_n=operatorname{ker}(d_n)$, and $frac{operatorname{ker}(d^*_n)}{operatorname{im}(d^*_{n+1})}=0$.
I need to show that $operatorname{ker}(d_n)subseteq operatorname{im}(d_{n+1})$. Let $zinker(d_n)$, then $operatorname{id_{Z_n}}inoperatorname{Hom}(Z_n,C_n)$ since $Z_nsubseteq C_n$. $operatorname{id_{Z_n}}inker(d^*_n)=operatorname{im}(d^*_{n+1})$, $z=operatorname{id_{Z_n}}(z)=d_{n+1}circ f_{n+1}(z)$ for some $f_{n+1}in operatorname{Hom}(Z_n,C_n)$, so $$ker(d_n)subseteqoperatorname{im}(d_{n+1})$$ So $H_n(C)=0$. Is this reasoning correct?
I'm also having trouble with the converse:
$operatorname{im}(d^*_{n+1})subseteqker(d^*_n)$ is OK- Let $f_ninker(d^*_n)$, then $d_ncirc f_n(z)=0$ and so $f_n(z)inker(d_n)=operatorname{im}(d_{n+1})$. Therefore $f_n(z)=d_{n+1}(y)$ for some $yin C_{n+1}$. I think I'm done if I can find a morphism $$g:Z_nrightarrow C_{n+1}$$ such that $g(z)=y$, but I'm having trouble arguing when this is possible (if it is).
Any help is much appreciated!
(Edit: $operatorname{id}_{Z_n}inoperatorname{ker}(d^*_n)$ is an arrow, and not an element)
Are there any criteria for when the diagram above commutes? For instance, by hypothesis $Z_ncong B_n=operatorname{im}(d_{n+1})$, so do I need $g$ to be surjective which would mean $g$ is the inverse of $d_{n+1}$?
category-theory homological-algebra
asked Jan 4 '18 at 17:38
cansomeonehelpmeoutcansomeonehelpmeout
6,8273835
$endgroup$
add a comment |
$begingroup$
I'm new to homological algebra, and I'm trying to learn it using Weibel's An introduction to homological algebra. I'm currently having trouble with exercise 1.1.4:
Show that ${operatorname{Hom}_R(A, C_n)}$ forms a chain complex of abelian groups for every $R$-module $A$ and every $R$-module chain complex $C$. Taking $A=Z_n$ show that if $H_n(operatorname{Hom}_R(Z_n,C))=0$, then $H_n(C)=0$. Is the converse true?
I think the first part is OK, using that if the chain complex looks like:
$$C:~ldotsrightarrow C_noverset{d_n}{rightarrow}C_{n-1}rightarrowldots$$
Then we get a new chain complex:
$$C^*:~ldotsrightarrowoperatorname{Hom}(A,C_n)overset{d^*_n}{rightarrow}operatorname{Hom}(A,C_{n-1})rightarrowldots$$
Where $d^*(f_n)=d_ncirc f_n$ for a $f_ninoperatorname{Hom}(A,C_n)$. This is a chain complex since $d^*_{n-1}circ d^*_n(f_n)=(d_{n-1}circ d_n)(f_n)=0$.
I'm not so sure about the second part. Assuming $Z_n=operatorname{ker}(d_n)$, and $frac{operatorname{ker}(d^*_n)}{operatorname{im}(d^*_{n+1})}=0$.
I need to show that $operatorname{ker}(d_n)subseteq operatorname{im}(d_{n+1})$. Let $zinker(d_n)$, then $operatorname{id_{Z_n}}inoperatorname{Hom}(Z_n,C_n)$ since $Z_nsubseteq C_n$. $operatorname{id_{Z_n}}inker(d^*_n)=operatorname{im}(d^*_{n+1})$, $z=operatorname{id_{Z_n}}(z)=d_{n+1}circ f_{n+1}(z)$ for some $f_{n+1}in operatorname{Hom}(Z_n,C_n)$, so $$ker(d_n)subseteqoperatorname{im}(d_{n+1})$$ So $H_n(C)=0$. Is this reasoning correct?
I'm also having trouble with the converse:
$operatorname{im}(d^*_{n+1})subseteqker(d^*_n)$ is OK- Let $f_ninker(d^*_n)$, then $d_ncirc f_n(z)=0$ and so $f_n(z)inker(d_n)=operatorname{im}(d_{n+1})$. Therefore $f_n(z)=d_{n+1}(y)$ for some $yin C_{n+1}$. I think I'm done if I can find a morphism $$g:Z_nrightarrow C_{n+1}$$ such that $g(z)=y$, but I'm having trouble arguing when this is possible (if it is).
Any help is much appreciated!
(Edit: $operatorname{id}_{Z_n}inoperatorname{ker}(d^*_n)$ is an arrow, and not an element)
Are there any criteria for when the diagram above commutes? For instance, by hypothesis $Z_ncong B_n=operatorname{im}(d_{n+1})$, so do I need $g$ to be surjective which would mean $g$ is the inverse of $d_{n+1}$?
category-theory homological-algebra
asked Jan 4 '18 at 17:38
cansomeonehelpmeoutcansomeonehelpmeout
6,8273835
$endgroup$
$begingroup$
Notice that $id_{Z_n}(z)$ is an element of $C_{n}$, not of $ker(d_n^*)$ (the latter is a set of arrows, not of elements). The hypothesis $H_n(Hom(Z_n,C_n))=0$ actually tells you that every arrow $f:Z_nrightarrow C_n$ such that $d_n f=0$ can be seen as a composition $d_{n+1}g$ for some other arrow $g:Z_nrightarrow C_{n+1}$: now you only need to take the right $f$ and you're done
$endgroup$
– TheMadcapLaughs
Jan 4 '18 at 22:02
$begingroup$
@TheMadcapLaughs in fact, the first object is fixed.
$endgroup$
– Rafael Holanda
Jan 5 '18 at 2:40
add a comment |
$begingroup$
I'm new to homological algebra, and I'm trying to learn it using Weibel's An introduction to homological algebra. I'm currently having trouble with exercise 1.1.4:
Show that ${operatorname{Hom}_R(A, C_n)}$ forms a chain complex of abelian groups for every $R$-module $A$ and every $R$-module chain complex $C$. Taking $A=Z_n$ show that if $H_n(operatorname{Hom}_R(Z_n,C))=0$, then $H_n(C)=0$. Is the converse true?
I think the first part is OK, using that if the chain complex looks like:
$$C:~ldotsrightarrow C_noverset{d_n}{rightarrow}C_{n-1}rightarrowldots$$
Then we get a new chain complex:
$$C^*:~ldotsrightarrowoperatorname{Hom}(A,C_n)overset{d^*_n}{rightarrow}operatorname{Hom}(A,C_{n-1})rightarrowldots$$
Where $d^*(f_n)=d_ncirc f_n$ for a $f_ninoperatorname{Hom}(A,C_n)$. This is a chain complex since $d^*_{n-1}circ d^*_n(f_n)=(d_{n-1}circ d_n)(f_n)=0$.
I'm not so sure about the second part. Assuming $Z_n=operatorname{ker}(d_n)$, and $frac{operatorname{ker}(d^*_n)}{operatorname{im}(d^*_{n+1})}=0$.
I need to show that $operatorname{ker}(d_n)subseteq operatorname{im}(d_{n+1})$. Let $zinker(d_n)$, then $operatorname{id_{Z_n}}inoperatorname{Hom}(Z_n,C_n)$ since $Z_nsubseteq C_n$. $operatorname{id_{Z_n}}inker(d^*_n)=operatorname{im}(d^*_{n+1})$, $z=operatorname{id_{Z_n}}(z)=d_{n+1}circ f_{n+1}(z)$ for some $f_{n+1}in operatorname{Hom}(Z_n,C_n)$, so $$ker(d_n)subseteqoperatorname{im}(d_{n+1})$$ So $H_n(C)=0$. Is this reasoning correct?
I'm also having trouble with the converse:
$operatorname{im}(d^*_{n+1})subseteqker(d^*_n)$ is OK- Let $f_ninker(d^*_n)$, then $d_ncirc f_n(z)=0$ and so $f_n(z)inker(d_n)=operatorname{im}(d_{n+1})$. Therefore $f_n(z)=d_{n+1}(y)$ for some $yin C_{n+1}$. I think I'm done if I can find a morphism $$g:Z_nrightarrow C_{n+1}$$ such that $g(z)=y$, but I'm having trouble arguing when this is possible (if it is).
Any help is much appreciated!
(Edit: $operatorname{id}_{Z_n}inoperatorname{ker}(d^*_n)$ is an arrow, and not an element)
Are there any criteria for when the diagram above commutes? For instance, by hypothesis $Z_ncong B_n=operatorname{im}(d_{n+1})$, so do I need $g$ to be surjective which would mean $g$ is the inverse of $d_{n+1}$?
category-theory homological-algebra
asked Jan 4 '18 at 17:38
cansomeonehelpmeoutcansomeonehelpmeout
6,8273835
$endgroup$
I'm new to homological algebra, and I'm trying to learn it using Weibel's An introduction to homological algebra. I'm currently having trouble with exercise 1.1.4:
Show that ${operatorname{Hom}_R(A, C_n)}$ forms a chain complex of abelian groups for every $R$-module $A$ and every $R$-module chain complex $C$. Taking $A=Z_n$ show that if $H_n(operatorname{Hom}_R(Z_n,C))=0$, then $H_n(C)=0$. Is the converse true?
I think the first part is OK, using that if the chain complex looks like:
$$C:~ldotsrightarrow C_noverset{d_n}{rightarrow}C_{n-1}rightarrowldots$$
Then we get a new chain complex:
$$C^*:~ldotsrightarrowoperatorname{Hom}(A,C_n)overset{d^*_n}{rightarrow}operatorname{Hom}(A,C_{n-1})rightarrowldots$$
Where $d^*(f_n)=d_ncirc f_n$ for a $f_ninoperatorname{Hom}(A,C_n)$. This is a chain complex since $d^*_{n-1}circ d^*_n(f_n)=(d_{n-1}circ d_n)(f_n)=0$.
I'm not so sure about the second part. Assuming $Z_n=operatorname{ker}(d_n)$, and $frac{operatorname{ker}(d^*_n)}{operatorname{im}(d^*_{n+1})}=0$.
I need to show that $operatorname{ker}(d_n)subseteq operatorname{im}(d_{n+1})$. Let $zinker(d_n)$, then $operatorname{id_{Z_n}}inoperatorname{Hom}(Z_n,C_n)$ since $Z_nsubseteq C_n$. $operatorname{id_{Z_n}}inker(d^*_n)=operatorname{im}(d^*_{n+1})$, $z=operatorname{id_{Z_n}}(z)=d_{n+1}circ f_{n+1}(z)$ for some $f_{n+1}in operatorname{Hom}(Z_n,C_n)$, so $$ker(d_n)subseteqoperatorname{im}(d_{n+1})$$ So $H_n(C)=0$. Is this reasoning correct?
I'm also having trouble with the converse:
$operatorname{im}(d^*_{n+1})subseteqker(d^*_n)$ is OK- Let $f_ninker(d^*_n)$, then $d_ncirc f_n(z)=0$ and so $f_n(z)inker(d_n)=operatorname{im}(d_{n+1})$. Therefore $f_n(z)=d_{n+1}(y)$ for some $yin C_{n+1}$. I think I'm done if I can find a morphism $$g:Z_nrightarrow C_{n+1}$$ such that $g(z)=y$, but I'm having trouble arguing when this is possible (if it is).
Any help is much appreciated!
(Edit: $operatorname{id}_{Z_n}inoperatorname{ker}(d^*_n)$ is an arrow, and not an element)
Are there any criteria for when the diagram above commutes? For instance, by hypothesis $Z_ncong B_n=operatorname{im}(d_{n+1})$, so do I need $g$ to be surjective which would mean $g$ is the inverse of $d_{n+1}$?
category-theory homological-algebra
category-theory homological-algebra
asked Jan 4 '18 at 17:38
cansomeonehelpmeoutcansomeonehelpmeout
6,8273835
asked Jan 4 '18 at 17:38
cansomeonehelpmeoutcansomeonehelpmeout
6,8273835
edited Dec 7 '18 at 19:50
cansomeonehelpmeout
asked Jan 4 '18 at 17:38
cansomeonehelpmeoutcansomeonehelpmeout
6,8273835
asked Jan 4 '18 at 17:38
cansomeonehelpmeoutcansomeonehelpmeout
6,8273835
asked Jan 4 '18 at 17:38
cansomeonehelpmeoutcansomeonehelpmeout
6,8273835
6,8273835
$begingroup$
Notice that $id_{Z_n}(z)$ is an element of $C_{n}$, not of $ker(d_n^*)$ (the latter is a set of arrows, not of elements). The hypothesis $H_n(Hom(Z_n,C_n))=0$ actually tells you that every arrow $f:Z_nrightarrow C_n$ such that $d_n f=0$ can be seen as a composition $d_{n+1}g$ for some other arrow $g:Z_nrightarrow C_{n+1}$: now you only need to take the right $f$ and you're done
$endgroup$
– TheMadcapLaughs
Jan 4 '18 at 22:02
$begingroup$
@TheMadcapLaughs in fact, the first object is fixed.
$endgroup$
– Rafael Holanda
Jan 5 '18 at 2:40
add a comment |
$begingroup$
Notice that $id_{Z_n}(z)$ is an element of $C_{n}$, not of $ker(d_n^*)$ (the latter is a set of arrows, not of elements). The hypothesis $H_n(Hom(Z_n,C_n))=0$ actually tells you that every arrow $f:Z_nrightarrow C_n$ such that $d_n f=0$ can be seen as a composition $d_{n+1}g$ for some other arrow $g:Z_nrightarrow C_{n+1}$: now you only need to take the right $f$ and you're done
$endgroup$
– TheMadcapLaughs
Jan 4 '18 at 22:02
$begingroup$
@TheMadcapLaughs in fact, the first object is fixed.
$endgroup$
– Rafael Holanda
Jan 5 '18 at 2:40
$begingroup$
Notice that $id_{Z_n}(z)$ is an element of $C_{n}$, not of $ker(d_n^*)$ (the latter is a set of arrows, not of elements). The hypothesis $H_n(Hom(Z_n,C_n))=0$ actually tells you that every arrow $f:Z_nrightarrow C_n$ such that $d_n f=0$ can be seen as a composition $d_{n+1}g$ for some other arrow $g:Z_nrightarrow C_{n+1}$: now you only need to take the right $f$ and you're done
$endgroup$
– TheMadcapLaughs
Jan 4 '18 at 22:02
$begingroup$
Notice that $id_{Z_n}(z)$ is an element of $C_{n}$, not of $ker(d_n^*)$ (the latter is a set of arrows, not of elements). The hypothesis $H_n(Hom(Z_n,C_n))=0$ actually tells you that every arrow $f:Z_nrightarrow C_n$ such that $d_n f=0$ can be seen as a composition $d_{n+1}g$ for some other arrow $g:Z_nrightarrow C_{n+1}$: now you only need to take the right $f$ and you're done
$endgroup$
– TheMadcapLaughs
Jan 4 '18 at 22:02
$begingroup$
@TheMadcapLaughs in fact, the first object is fixed.
$endgroup$
– Rafael Holanda
Jan 5 '18 at 2:40
$begingroup$
@TheMadcapLaughs in fact, the first object is fixed.
$endgroup$
– Rafael Holanda
Jan 5 '18 at 2:40
add a comment |
1 Answer
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$begingroup$
Elements of $ker(d_n^*)$ are arrows. Let $i_n:Z_nrightarrow C_n$ be the canonical inclusion. Then $i_nin Hom(Z_n,C_n)$ and
$$d^*_n(i_n)=d_ncirc i_n=0.$$
Hence there exists $uin Hom(Z_n,C_{n+1})$ such that $i_n=d_{n+1}circ u$ and therefore
$$Z_nsubseteq B_n=im(d_{n+1})$$
Note that this argument is valid to any exact category.
The reciproque is false: consider the complex $0rightarrow 2mathbb{Z}rightarrowmathbb{Z}rightarrowmathbb{Z}_2rightarrow0$ and take $Z_n=mathbb{Z}_2$.
answered Jan 5 '18 at 2:39
Rafael HolandaRafael Holanda
2,706722
$endgroup$
2
$begingroup$
Thank you for the counterexample! So it is not true because I would get a new complex: $0rightarrow 0rightarrow 0rightarrow mathbb{Z}_2rightarrow 0$, and $frac{operatorname{ker}(mathbb{Z_2}rightarrow 0)}{operatorname{im}(0rightarrowmathbb{Z}_2)}congmathbb{Z}_2neq 0$? I was also looking for when the converse holds (I've updated the question).
$endgroup$
– cansomeonehelpmeout
Jan 10 '18 at 22:06
add a comment |
Your Answer
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$begingroup$
Elements of $ker(d_n^*)$ are arrows. Let $i_n:Z_nrightarrow C_n$ be the canonical inclusion. Then $i_nin Hom(Z_n,C_n)$ and
$$d^*_n(i_n)=d_ncirc i_n=0.$$
Hence there exists $uin Hom(Z_n,C_{n+1})$ such that $i_n=d_{n+1}circ u$ and therefore
$$Z_nsubseteq B_n=im(d_{n+1})$$
Note that this argument is valid to any exact category.
The reciproque is false: consider the complex $0rightarrow 2mathbb{Z}rightarrowmathbb{Z}rightarrowmathbb{Z}_2rightarrow0$ and take $Z_n=mathbb{Z}_2$.
answered Jan 5 '18 at 2:39
Rafael HolandaRafael Holanda
2,706722
$endgroup$
2
$begingroup$
Thank you for the counterexample! So it is not true because I would get a new complex: $0rightarrow 0rightarrow 0rightarrow mathbb{Z}_2rightarrow 0$, and $frac{operatorname{ker}(mathbb{Z_2}rightarrow 0)}{operatorname{im}(0rightarrowmathbb{Z}_2)}congmathbb{Z}_2neq 0$? I was also looking for when the converse holds (I've updated the question).
$endgroup$
– cansomeonehelpmeout
Jan 10 '18 at 22:06
add a comment |
$begingroup$
Elements of $ker(d_n^*)$ are arrows. Let $i_n:Z_nrightarrow C_n$ be the canonical inclusion. Then $i_nin Hom(Z_n,C_n)$ and
$$d^*_n(i_n)=d_ncirc i_n=0.$$
Hence there exists $uin Hom(Z_n,C_{n+1})$ such that $i_n=d_{n+1}circ u$ and therefore
$$Z_nsubseteq B_n=im(d_{n+1})$$
Note that this argument is valid to any exact category.
The reciproque is false: consider the complex $0rightarrow 2mathbb{Z}rightarrowmathbb{Z}rightarrowmathbb{Z}_2rightarrow0$ and take $Z_n=mathbb{Z}_2$.
answered Jan 5 '18 at 2:39
Rafael HolandaRafael Holanda
2,706722
$endgroup$
2
$begingroup$
Thank you for the counterexample! So it is not true because I would get a new complex: $0rightarrow 0rightarrow 0rightarrow mathbb{Z}_2rightarrow 0$, and $frac{operatorname{ker}(mathbb{Z_2}rightarrow 0)}{operatorname{im}(0rightarrowmathbb{Z}_2)}congmathbb{Z}_2neq 0$? I was also looking for when the converse holds (I've updated the question).
$endgroup$
– cansomeonehelpmeout
Jan 10 '18 at 22:06
add a comment |
$begingroup$
Elements of $ker(d_n^*)$ are arrows. Let $i_n:Z_nrightarrow C_n$ be the canonical inclusion. Then $i_nin Hom(Z_n,C_n)$ and
$$d^*_n(i_n)=d_ncirc i_n=0.$$
Hence there exists $uin Hom(Z_n,C_{n+1})$ such that $i_n=d_{n+1}circ u$ and therefore
$$Z_nsubseteq B_n=im(d_{n+1})$$
Note that this argument is valid to any exact category.
The reciproque is false: consider the complex $0rightarrow 2mathbb{Z}rightarrowmathbb{Z}rightarrowmathbb{Z}_2rightarrow0$ and take $Z_n=mathbb{Z}_2$.
answered Jan 5 '18 at 2:39
Rafael HolandaRafael Holanda
2,706722
$endgroup$
Elements of $ker(d_n^*)$ are arrows. Let $i_n:Z_nrightarrow C_n$ be the canonical inclusion. Then $i_nin Hom(Z_n,C_n)$ and
$$d^*_n(i_n)=d_ncirc i_n=0.$$
Hence there exists $uin Hom(Z_n,C_{n+1})$ such that $i_n=d_{n+1}circ u$ and therefore
$$Z_nsubseteq B_n=im(d_{n+1})$$
Note that this argument is valid to any exact category.
The reciproque is false: consider the complex $0rightarrow 2mathbb{Z}rightarrowmathbb{Z}rightarrowmathbb{Z}_2rightarrow0$ and take $Z_n=mathbb{Z}_2$.
answered Jan 5 '18 at 2:39
Rafael HolandaRafael Holanda
2,706722
answered Jan 5 '18 at 2:39
Rafael HolandaRafael Holanda
2,706722
answered Jan 5 '18 at 2:39
Rafael HolandaRafael Holanda
2,706722
answered Jan 5 '18 at 2:39
Rafael HolandaRafael Holanda
2,706722
2,706722
2
$begingroup$
Thank you for the counterexample! So it is not true because I would get a new complex: $0rightarrow 0rightarrow 0rightarrow mathbb{Z}_2rightarrow 0$, and $frac{operatorname{ker}(mathbb{Z_2}rightarrow 0)}{operatorname{im}(0rightarrowmathbb{Z}_2)}congmathbb{Z}_2neq 0$? I was also looking for when the converse holds (I've updated the question).
$endgroup$
– cansomeonehelpmeout
Jan 10 '18 at 22:06
add a comment |
2
$begingroup$
Thank you for the counterexample! So it is not true because I would get a new complex: $0rightarrow 0rightarrow 0rightarrow mathbb{Z}_2rightarrow 0$, and $frac{operatorname{ker}(mathbb{Z_2}rightarrow 0)}{operatorname{im}(0rightarrowmathbb{Z}_2)}congmathbb{Z}_2neq 0$? I was also looking for when the converse holds (I've updated the question).
$endgroup$
– cansomeonehelpmeout
Jan 10 '18 at 22:06
2
2
$begingroup$
Thank you for the counterexample! So it is not true because I would get a new complex: $0rightarrow 0rightarrow 0rightarrow mathbb{Z}_2rightarrow 0$, and $frac{operatorname{ker}(mathbb{Z_2}rightarrow 0)}{operatorname{im}(0rightarrowmathbb{Z}_2)}congmathbb{Z}_2neq 0$? I was also looking for when the converse holds (I've updated the question).
$endgroup$
– cansomeonehelpmeout
Jan 10 '18 at 22:06
$begingroup$
Thank you for the counterexample! So it is not true because I would get a new complex: $0rightarrow 0rightarrow 0rightarrow mathbb{Z}_2rightarrow 0$, and $frac{operatorname{ker}(mathbb{Z_2}rightarrow 0)}{operatorname{im}(0rightarrowmathbb{Z}_2)}congmathbb{Z}_2neq 0$? I was also looking for when the converse holds (I've updated the question).
$endgroup$
– cansomeonehelpmeout
Jan 10 '18 at 22:06
add a comment |
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Notice that $id_{Z_n}(z)$ is an element of $C_{n}$, not of $ker(d_n^*)$ (the latter is a set of arrows, not of elements). The hypothesis $H_n(Hom(Z_n,C_n))=0$ actually tells you that every arrow $f:Z_nrightarrow C_n$ such that $d_n f=0$ can be seen as a composition $d_{n+1}g$ for some other arrow $g:Z_nrightarrow C_{n+1}$: now you only need to take the right $f$ and you're done
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– TheMadcapLaughs
Jan 4 '18 at 22:02
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@TheMadcapLaughs in fact, the first object is fixed.
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– Rafael Holanda
Jan 5 '18 at 2:40