Mixing
Proving “if $A,B$ are groups, $f: A mapsto B$ a homomorphism, $S leqslant B$, then $f^{-1}(S) leqslant...
The following theorem is an exercise in a textbook:
Theorem: Let $(A, cdot)$, $(B, ast)$ be groups, let $f: A mapsto B$ be a homomorphism, and let $S leqslant B$. Then:
$$textrm{Ker}(f) leqslant f^{-1}(S) = {x in G mid f(x) in S} leqslant A $$
I provided a proof for it here, but I am uncomfortable with this proof.
Shouldn't the theorem also provide some surjectivity conditions on $f$? At least, there should exist some $S leqslant B$, such that for all $y in S$, there exists $x in A$ such that $f(x) = y$?
Say that this condition does not hold, so that there exists $p in S$ such that for all $x in A$, $f(x) neq y$. Assume further that there are elements in $f(a), f(b) in S$, which are in the image of $f$, such that $f(a) ast f(b) = p$. Then, $f^{-1}(S)$ would not be closed, since $p = f(a) ast f(b) = f(a cdot b) notin A$, so it cannot be that $f^{-1}(S) leqslant A$?
group-theory proof-explanation
asked Nov 22 '18 at 1:06
user89user89
6861647
add a comment |
The following theorem is an exercise in a textbook:
Theorem: Let $(A, cdot)$, $(B, ast)$ be groups, let $f: A mapsto B$ be a homomorphism, and let $S leqslant B$. Then:
$$textrm{Ker}(f) leqslant f^{-1}(S) = {x in G mid f(x) in S} leqslant A $$
I provided a proof for it here, but I am uncomfortable with this proof.
Shouldn't the theorem also provide some surjectivity conditions on $f$? At least, there should exist some $S leqslant B$, such that for all $y in S$, there exists $x in A$ such that $f(x) = y$?
Say that this condition does not hold, so that there exists $p in S$ such that for all $x in A$, $f(x) neq y$. Assume further that there are elements in $f(a), f(b) in S$, which are in the image of $f$, such that $f(a) ast f(b) = p$. Then, $f^{-1}(S)$ would not be closed, since $p = f(a) ast f(b) = f(a cdot b) notin A$, so it cannot be that $f^{-1}(S) leqslant A$?
group-theory proof-explanation
asked Nov 22 '18 at 1:06
user89user89
6861647
add a comment |
The following theorem is an exercise in a textbook:
Theorem: Let $(A, cdot)$, $(B, ast)$ be groups, let $f: A mapsto B$ be a homomorphism, and let $S leqslant B$. Then:
$$textrm{Ker}(f) leqslant f^{-1}(S) = {x in G mid f(x) in S} leqslant A $$
I provided a proof for it here, but I am uncomfortable with this proof.
Shouldn't the theorem also provide some surjectivity conditions on $f$? At least, there should exist some $S leqslant B$, such that for all $y in S$, there exists $x in A$ such that $f(x) = y$?
Say that this condition does not hold, so that there exists $p in S$ such that for all $x in A$, $f(x) neq y$. Assume further that there are elements in $f(a), f(b) in S$, which are in the image of $f$, such that $f(a) ast f(b) = p$. Then, $f^{-1}(S)$ would not be closed, since $p = f(a) ast f(b) = f(a cdot b) notin A$, so it cannot be that $f^{-1}(S) leqslant A$?
group-theory proof-explanation
asked Nov 22 '18 at 1:06
user89user89
6861647
The following theorem is an exercise in a textbook:
Theorem: Let $(A, cdot)$, $(B, ast)$ be groups, let $f: A mapsto B$ be a homomorphism, and let $S leqslant B$. Then:
$$textrm{Ker}(f) leqslant f^{-1}(S) = {x in G mid f(x) in S} leqslant A $$
I provided a proof for it here, but I am uncomfortable with this proof.
Shouldn't the theorem also provide some surjectivity conditions on $f$? At least, there should exist some $S leqslant B$, such that for all $y in S$, there exists $x in A$ such that $f(x) = y$?
Say that this condition does not hold, so that there exists $p in S$ such that for all $x in A$, $f(x) neq y$. Assume further that there are elements in $f(a), f(b) in S$, which are in the image of $f$, such that $f(a) ast f(b) = p$. Then, $f^{-1}(S)$ would not be closed, since $p = f(a) ast f(b) = f(a cdot b) notin A$, so it cannot be that $f^{-1}(S) leqslant A$?
group-theory proof-explanation
group-theory proof-explanation
asked Nov 22 '18 at 1:06
user89user89
6861647
asked Nov 22 '18 at 1:06
user89user89
6861647
asked Nov 22 '18 at 1:06
user89user89
6861647
asked Nov 22 '18 at 1:06
user89user89
6861647
asked Nov 22 '18 at 1:06
user89user89
6861647
6861647
add a comment |
add a comment |
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