Mixing
Restricting domain of surjective function to make bijection
up vote
2
down vote
favorite
Statement:
For every surjective function $f:A to B$ there exists set $Csubseteq A$ such that function $f:C to B$ is bijection.
As I see it, this is obviously true for finite sets, in way that for every multiple occurrence of some element in $B$, it is possible to just eliminate all elements in $A$ whose image is that specific element in $B$ but one. However I am not sure about infinite sets, as I can't quite come up with counterexample, if there is one.
Thanks for help.
functions
edited yesterday
N. F. Taussig
42.4k93254
asked yesterday
Dovla
588
add a comment |
up vote
2
down vote
favorite
Statement:
For every surjective function $f:A to B$ there exists set $Csubseteq A$ such that function $f:C to B$ is bijection.
As I see it, this is obviously true for finite sets, in way that for every multiple occurrence of some element in $B$, it is possible to just eliminate all elements in $A$ whose image is that specific element in $B$ but one. However I am not sure about infinite sets, as I can't quite come up with counterexample, if there is one.
Thanks for help.
functions
edited yesterday
N. F. Taussig
42.4k93254
asked yesterday
Dovla
588
2
There's nothing in your argument that depends on anything being finite (with the possible exception of dependience on the Axiom of Choice, which you need pay no attention to at this stage of your studies).
– Ethan Bolker
yesterday
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Statement:
For every surjective function $f:A to B$ there exists set $Csubseteq A$ such that function $f:C to B$ is bijection.
As I see it, this is obviously true for finite sets, in way that for every multiple occurrence of some element in $B$, it is possible to just eliminate all elements in $A$ whose image is that specific element in $B$ but one. However I am not sure about infinite sets, as I can't quite come up with counterexample, if there is one.
Thanks for help.
functions
edited yesterday
N. F. Taussig
42.4k93254
asked yesterday
Dovla
588
Statement:
For every surjective function $f:A to B$ there exists set $Csubseteq A$ such that function $f:C to B$ is bijection.
As I see it, this is obviously true for finite sets, in way that for every multiple occurrence of some element in $B$, it is possible to just eliminate all elements in $A$ whose image is that specific element in $B$ but one. However I am not sure about infinite sets, as I can't quite come up with counterexample, if there is one.
Thanks for help.
functions
functions
edited yesterday
N. F. Taussig
42.4k93254
asked yesterday
Dovla
588
edited yesterday
N. F. Taussig
42.4k93254
asked yesterday
Dovla
588
edited yesterday
N. F. Taussig
42.4k93254
edited yesterday
N. F. Taussig
42.4k93254
edited yesterday
N. F. Taussig
42.4k93254
42.4k93254
asked yesterday
Dovla
588
asked yesterday
Dovla
588
asked yesterday
Dovla
588
588
2
There's nothing in your argument that depends on anything being finite (with the possible exception of dependience on the Axiom of Choice, which you need pay no attention to at this stage of your studies).
– Ethan Bolker
yesterday
add a comment |
2
There's nothing in your argument that depends on anything being finite (with the possible exception of dependience on the Axiom of Choice, which you need pay no attention to at this stage of your studies).
– Ethan Bolker
yesterday
2
2
There's nothing in your argument that depends on anything being finite (with the possible exception of dependience on the Axiom of Choice, which you need pay no attention to at this stage of your studies).
– Ethan Bolker
yesterday
There's nothing in your argument that depends on anything being finite (with the possible exception of dependience on the Axiom of Choice, which you need pay no attention to at this stage of your studies).
– Ethan Bolker
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
If $f:Arightarrow B$ is a surjective mapping, then the set of preimages ${f^{-1}(b)mid bin B}$ with $f^{-1}(b) = {ain Amid f(a)=b}$ forms a partition of $A$. Now choose for each set $f^{-1}(b)$ one representative $a_bin A$ with $f(a_b)=b$ and you are done. See Bolker's comment.
answered yesterday
Wuestenfux
2,3441410
1
+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $f:Arightarrow B$ is a surjective mapping, then the set of preimages ${f^{-1}(b)mid bin B}$ with $f^{-1}(b) = {ain Amid f(a)=b}$ forms a partition of $A$. Now choose for each set $f^{-1}(b)$ one representative $a_bin A$ with $f(a_b)=b$ and you are done. See Bolker's comment.
answered yesterday
Wuestenfux
2,3441410
1
+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday
add a comment |
up vote
2
down vote
If $f:Arightarrow B$ is a surjective mapping, then the set of preimages ${f^{-1}(b)mid bin B}$ with $f^{-1}(b) = {ain Amid f(a)=b}$ forms a partition of $A$. Now choose for each set $f^{-1}(b)$ one representative $a_bin A$ with $f(a_b)=b$ and you are done. See Bolker's comment.
answered yesterday
Wuestenfux
2,3441410
1
+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
If $f:Arightarrow B$ is a surjective mapping, then the set of preimages ${f^{-1}(b)mid bin B}$ with $f^{-1}(b) = {ain Amid f(a)=b}$ forms a partition of $A$. Now choose for each set $f^{-1}(b)$ one representative $a_bin A$ with $f(a_b)=b$ and you are done. See Bolker's comment.
answered yesterday
Wuestenfux
2,3441410
If $f:Arightarrow B$ is a surjective mapping, then the set of preimages ${f^{-1}(b)mid bin B}$ with $f^{-1}(b) = {ain Amid f(a)=b}$ forms a partition of $A$. Now choose for each set $f^{-1}(b)$ one representative $a_bin A$ with $f(a_b)=b$ and you are done. See Bolker's comment.
answered yesterday
Wuestenfux
2,3441410
answered yesterday
Wuestenfux
2,3441410
answered yesterday
Wuestenfux
2,3441410
answered yesterday
Wuestenfux
2,3441410
2,3441410
1
+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday
add a comment |
1
+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday
1
1
+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday
+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004894%2frestricting-domain-of-surjective-function-to-make-bijection%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
There's nothing in your argument that depends on anything being finite (with the possible exception of dependience on the Axiom of Choice, which you need pay no attention to at this stage of your studies).
– Ethan Bolker
yesterday