How to check if a function is injective and surjective [closed]
up vote
0
down vote
favorite
I'm currentlly doing a course in abstract algebra and I often have to prove a map is surjective or injective. It's always done the same way, we take $f(a)=f(b)$ and deduce $a=b$, or we show that for every $y$ in the range there is an element x in the domain such that $f(x)=y$.
I was wondering if there are alternative ways we can use to prove a map is injective/surjective?
alternative-proof
asked Nov 17 at 19:02
Lowkey
467
closed as too broad by Yanko, Shailesh, Leucippus, user10354138, Parcly Taxel Nov 18 at 2:53
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
I'm currentlly doing a course in abstract algebra and I often have to prove a map is surjective or injective. It's always done the same way, we take $f(a)=f(b)$ and deduce $a=b$, or we show that for every $y$ in the range there is an element x in the domain such that $f(x)=y$.
I was wondering if there are alternative ways we can use to prove a map is injective/surjective?
alternative-proof
asked Nov 17 at 19:02
Lowkey
467
closed as too broad by Yanko, Shailesh, Leucippus, user10354138, Parcly Taxel Nov 18 at 2:53
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
You can prove that it is bijective by demonstrating an inverse.
– Patrick Stevens
Nov 17 at 19:04
You can try to show that $f$ is a composition of some injective/surjective functions. Anyway I vote to close this as being too broad.
– Yanko
Nov 17 at 19:05
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm currentlly doing a course in abstract algebra and I often have to prove a map is surjective or injective. It's always done the same way, we take $f(a)=f(b)$ and deduce $a=b$, or we show that for every $y$ in the range there is an element x in the domain such that $f(x)=y$.
I was wondering if there are alternative ways we can use to prove a map is injective/surjective?
alternative-proof
asked Nov 17 at 19:02
Lowkey
467
I'm currentlly doing a course in abstract algebra and I often have to prove a map is surjective or injective. It's always done the same way, we take $f(a)=f(b)$ and deduce $a=b$, or we show that for every $y$ in the range there is an element x in the domain such that $f(x)=y$.
I was wondering if there are alternative ways we can use to prove a map is injective/surjective?
alternative-proof
alternative-proof
asked Nov 17 at 19:02
Lowkey
467
asked Nov 17 at 19:02
Lowkey
467
asked Nov 17 at 19:02
Lowkey
467
asked Nov 17 at 19:02
Lowkey
467
asked Nov 17 at 19:02
Lowkey
467
467
closed as too broad by Yanko, Shailesh, Leucippus, user10354138, Parcly Taxel Nov 18 at 2:53
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by Yanko, Shailesh, Leucippus, user10354138, Parcly Taxel Nov 18 at 2:53
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
You can prove that it is bijective by demonstrating an inverse.
– Patrick Stevens
Nov 17 at 19:04
You can try to show that $f$ is a composition of some injective/surjective functions. Anyway I vote to close this as being too broad.
– Yanko
Nov 17 at 19:05
add a comment |
3
You can prove that it is bijective by demonstrating an inverse.
– Patrick Stevens
Nov 17 at 19:04
You can try to show that $f$ is a composition of some injective/surjective functions. Anyway I vote to close this as being too broad.
– Yanko
Nov 17 at 19:05
3
3
You can prove that it is bijective by demonstrating an inverse.
– Patrick Stevens
Nov 17 at 19:04
You can prove that it is bijective by demonstrating an inverse.
– Patrick Stevens
Nov 17 at 19:04
You can try to show that $f$ is a composition of some injective/surjective functions. Anyway I vote to close this as being too broad.
– Yanko
Nov 17 at 19:05
You can try to show that $f$ is a composition of some injective/surjective functions. Anyway I vote to close this as being too broad.
– Yanko
Nov 17 at 19:05
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Suppose that $X,Y,Z$ are sets and $f: X to Y$ and $g : Y to Z$ are functions. We have the following implications: if $g circ f$ is a surjection, then $g$ is surjection. If $g circ f$ is an injection, then $f$ is an injection. I encourage you to prove this lemma yourself if you haven't seen it before.
answered Nov 17 at 19:09
Blazej
1,474618
This was very helpful, thank you. I vaguely remembered it from class but I knew there was something similar to this.
– Lowkey
Nov 17 at 19:54
add a comment |
up vote
1
down vote
If the map is an arbitrary function, there is no better way.
However, if the map is a homomorphism, you can prove it is injective by showing its kernel is trivial.
answered Nov 17 at 19:03
helper
523212
1
I don't agree that this is the only way (see my answer). However the comment about checking kernel is very useful in practice. Similarly surjectivity may sometimes be checked by computing the cokernel. Be warned though that cokernels don't always exist (it depends on the type of algebraic structures you are working with).
– Blazej
Nov 17 at 19:11
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Suppose that $X,Y,Z$ are sets and $f: X to Y$ and $g : Y to Z$ are functions. We have the following implications: if $g circ f$ is a surjection, then $g$ is surjection. If $g circ f$ is an injection, then $f$ is an injection. I encourage you to prove this lemma yourself if you haven't seen it before.
answered Nov 17 at 19:09
Blazej
1,474618
This was very helpful, thank you. I vaguely remembered it from class but I knew there was something similar to this.
– Lowkey
Nov 17 at 19:54
add a comment |
up vote
0
down vote
accepted
Suppose that $X,Y,Z$ are sets and $f: X to Y$ and $g : Y to Z$ are functions. We have the following implications: if $g circ f$ is a surjection, then $g$ is surjection. If $g circ f$ is an injection, then $f$ is an injection. I encourage you to prove this lemma yourself if you haven't seen it before.
answered Nov 17 at 19:09
Blazej
1,474618
This was very helpful, thank you. I vaguely remembered it from class but I knew there was something similar to this.
– Lowkey
Nov 17 at 19:54
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Suppose that $X,Y,Z$ are sets and $f: X to Y$ and $g : Y to Z$ are functions. We have the following implications: if $g circ f$ is a surjection, then $g$ is surjection. If $g circ f$ is an injection, then $f$ is an injection. I encourage you to prove this lemma yourself if you haven't seen it before.
answered Nov 17 at 19:09
Blazej
1,474618
Suppose that $X,Y,Z$ are sets and $f: X to Y$ and $g : Y to Z$ are functions. We have the following implications: if $g circ f$ is a surjection, then $g$ is surjection. If $g circ f$ is an injection, then $f$ is an injection. I encourage you to prove this lemma yourself if you haven't seen it before.
answered Nov 17 at 19:09
Blazej
1,474618
answered Nov 17 at 19:09
Blazej
1,474618
answered Nov 17 at 19:09
Blazej
1,474618
answered Nov 17 at 19:09
Blazej
1,474618
1,474618
This was very helpful, thank you. I vaguely remembered it from class but I knew there was something similar to this.
– Lowkey
Nov 17 at 19:54
add a comment |
This was very helpful, thank you. I vaguely remembered it from class but I knew there was something similar to this.
– Lowkey
Nov 17 at 19:54
This was very helpful, thank you. I vaguely remembered it from class but I knew there was something similar to this.
– Lowkey
Nov 17 at 19:54
This was very helpful, thank you. I vaguely remembered it from class but I knew there was something similar to this.
– Lowkey
Nov 17 at 19:54
add a comment |
up vote
1
down vote
If the map is an arbitrary function, there is no better way.
However, if the map is a homomorphism, you can prove it is injective by showing its kernel is trivial.
answered Nov 17 at 19:03
helper
523212
1
I don't agree that this is the only way (see my answer). However the comment about checking kernel is very useful in practice. Similarly surjectivity may sometimes be checked by computing the cokernel. Be warned though that cokernels don't always exist (it depends on the type of algebraic structures you are working with).
– Blazej
Nov 17 at 19:11
add a comment |
up vote
1
down vote
If the map is an arbitrary function, there is no better way.
However, if the map is a homomorphism, you can prove it is injective by showing its kernel is trivial.
answered Nov 17 at 19:03
helper
523212
1
I don't agree that this is the only way (see my answer). However the comment about checking kernel is very useful in practice. Similarly surjectivity may sometimes be checked by computing the cokernel. Be warned though that cokernels don't always exist (it depends on the type of algebraic structures you are working with).
– Blazej
Nov 17 at 19:11
add a comment |
up vote
1
down vote
up vote
1
down vote
If the map is an arbitrary function, there is no better way.
However, if the map is a homomorphism, you can prove it is injective by showing its kernel is trivial.
answered Nov 17 at 19:03
helper
523212
If the map is an arbitrary function, there is no better way.
However, if the map is a homomorphism, you can prove it is injective by showing its kernel is trivial.
answered Nov 17 at 19:03
helper
523212
answered Nov 17 at 19:03
helper
523212
answered Nov 17 at 19:03
helper
523212
answered Nov 17 at 19:03
helper
523212
523212
1
I don't agree that this is the only way (see my answer). However the comment about checking kernel is very useful in practice. Similarly surjectivity may sometimes be checked by computing the cokernel. Be warned though that cokernels don't always exist (it depends on the type of algebraic structures you are working with).
– Blazej
Nov 17 at 19:11
add a comment |
1
I don't agree that this is the only way (see my answer). However the comment about checking kernel is very useful in practice. Similarly surjectivity may sometimes be checked by computing the cokernel. Be warned though that cokernels don't always exist (it depends on the type of algebraic structures you are working with).
– Blazej
Nov 17 at 19:11
1
1
I don't agree that this is the only way (see my answer). However the comment about checking kernel is very useful in practice. Similarly surjectivity may sometimes be checked by computing the cokernel. Be warned though that cokernels don't always exist (it depends on the type of algebraic structures you are working with).
– Blazej
Nov 17 at 19:11
I don't agree that this is the only way (see my answer). However the comment about checking kernel is very useful in practice. Similarly surjectivity may sometimes be checked by computing the cokernel. Be warned though that cokernels don't always exist (it depends on the type of algebraic structures you are working with).
– Blazej
Nov 17 at 19:11
add a comment |
3
You can prove that it is bijective by demonstrating an inverse.
– Patrick Stevens
Nov 17 at 19:04
You can try to show that $f$ is a composition of some injective/surjective functions. Anyway I vote to close this as being too broad.
– Yanko
Nov 17 at 19:05