Seeking examples of $f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$
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Could you give me some examples for this statement?
$$f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$$
general-topology
edited Nov 24 at 8:31
Blue
47.1k870148
asked Nov 24 at 8:19
Minh
1788
add a comment |
up vote
-1
down vote
favorite
Could you give me some examples for this statement?
$$f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$$
general-topology
edited Nov 24 at 8:31
Blue
47.1k870148
asked Nov 24 at 8:19
Minh
1788
1
the preimage of the codomain is the domain itself by definition
– Alvin Lepik
Nov 24 at 8:26
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Could you give me some examples for this statement?
$$f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$$
general-topology
edited Nov 24 at 8:31
Blue
47.1k870148
asked Nov 24 at 8:19
Minh
1788
Could you give me some examples for this statement?
$$f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$$
general-topology
general-topology
edited Nov 24 at 8:31
Blue
47.1k870148
asked Nov 24 at 8:19
Minh
1788
edited Nov 24 at 8:31
Blue
47.1k870148
asked Nov 24 at 8:19
Minh
1788
edited Nov 24 at 8:31
Blue
47.1k870148
edited Nov 24 at 8:31
Blue
47.1k870148
edited Nov 24 at 8:31
Blue
47.1k870148
47.1k870148
asked Nov 24 at 8:19
Minh
1788
asked Nov 24 at 8:19
Minh
1788
asked Nov 24 at 8:19
Minh
1788
1788
1
the preimage of the codomain is the domain itself by definition
– Alvin Lepik
Nov 24 at 8:26
add a comment |
1
the preimage of the codomain is the domain itself by definition
– Alvin Lepik
Nov 24 at 8:26
1
1
the preimage of the codomain is the domain itself by definition
– Alvin Lepik
Nov 24 at 8:26
the preimage of the codomain is the domain itself by definition
– Alvin Lepik
Nov 24 at 8:26
add a comment |
1 Answer
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3
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By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.
answered Nov 24 at 8:25
Henno Brandsma
102k345111
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.
answered Nov 24 at 8:25
Henno Brandsma
102k345111
add a comment |
up vote
3
down vote
By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.
answered Nov 24 at 8:25
Henno Brandsma
102k345111
add a comment |
up vote
3
down vote
up vote
3
down vote
By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.
answered Nov 24 at 8:25
Henno Brandsma
102k345111
By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.
answered Nov 24 at 8:25
Henno Brandsma
102k345111
edited Nov 24 at 14:16
answered Nov 24 at 8:25
Henno Brandsma
102k345111
answered Nov 24 at 8:25
Henno Brandsma
102k345111
answered Nov 24 at 8:25
Henno Brandsma
102k345111
102k345111
add a comment |
add a comment |
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the preimage of the codomain is the domain itself by definition
– Alvin Lepik
Nov 24 at 8:26