Prove $p_kcirc f$ continuous $implies$ f is continuous
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Let $X_1,dots X_n$ topological space and $p_k:X_1timescdots X_nto X_k$ the projection to the kth component. Let $Y$ be topological space and
$f:Yto X_1timescdotstimes X_n$ function s.t $forall 1le kle nquad p_kcirc f$ is continuous. Prove that $f$ is continuous.
I thought that since $p_kcirc f$ is continuous $forall1le kle n$, $(p_kcirc f)^{-1}(O_k)intau_{Y}$ (where $O_kintau_k$ the topology of $(X_k,tau_k)$) and then each open set in $tau_{pi}$ is a product of open sets from these topologies, i.e $$forall Ointau_pi,O=bigcup_{iin I}prod_{k=1}^n O_{k,i}$$ (where $O_{k,i}$ is an open set in $tau_k$)$$Rightarrow f^{-1}(O)intau_Y$$ but that's seems fishy to me. Am I (even not completely) right?
general-topology continuity
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add a comment |
$begingroup$
Let $X_1,dots X_n$ topological space and $p_k:X_1timescdots X_nto X_k$ the projection to the kth component. Let $Y$ be topological space and
$f:Yto X_1timescdotstimes X_n$ function s.t $forall 1le kle nquad p_kcirc f$ is continuous. Prove that $f$ is continuous.
I thought that since $p_kcirc f$ is continuous $forall1le kle n$, $(p_kcirc f)^{-1}(O_k)intau_{Y}$ (where $O_kintau_k$ the topology of $(X_k,tau_k)$) and then each open set in $tau_{pi}$ is a product of open sets from these topologies, i.e $$forall Ointau_pi,O=bigcup_{iin I}prod_{k=1}^n O_{k,i}$$ (where $O_{k,i}$ is an open set in $tau_k$)$$Rightarrow f^{-1}(O)intau_Y$$ but that's seems fishy to me. Am I (even not completely) right?
general-topology continuity
$endgroup$
$begingroup$
Each open set in the product topology is the union of products of open sets. So you're close to the target.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 10:33
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I added union before the product. Now that's correct?
$endgroup$
– user65985
Sep 7 '14 at 10:36
$begingroup$
You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 11:48
add a comment |
$begingroup$
Let $X_1,dots X_n$ topological space and $p_k:X_1timescdots X_nto X_k$ the projection to the kth component. Let $Y$ be topological space and
$f:Yto X_1timescdotstimes X_n$ function s.t $forall 1le kle nquad p_kcirc f$ is continuous. Prove that $f$ is continuous.
I thought that since $p_kcirc f$ is continuous $forall1le kle n$, $(p_kcirc f)^{-1}(O_k)intau_{Y}$ (where $O_kintau_k$ the topology of $(X_k,tau_k)$) and then each open set in $tau_{pi}$ is a product of open sets from these topologies, i.e $$forall Ointau_pi,O=bigcup_{iin I}prod_{k=1}^n O_{k,i}$$ (where $O_{k,i}$ is an open set in $tau_k$)$$Rightarrow f^{-1}(O)intau_Y$$ but that's seems fishy to me. Am I (even not completely) right?
general-topology continuity
$endgroup$
Let $X_1,dots X_n$ topological space and $p_k:X_1timescdots X_nto X_k$ the projection to the kth component. Let $Y$ be topological space and
$f:Yto X_1timescdotstimes X_n$ function s.t $forall 1le kle nquad p_kcirc f$ is continuous. Prove that $f$ is continuous.
I thought that since $p_kcirc f$ is continuous $forall1le kle n$, $(p_kcirc f)^{-1}(O_k)intau_{Y}$ (where $O_kintau_k$ the topology of $(X_k,tau_k)$) and then each open set in $tau_{pi}$ is a product of open sets from these topologies, i.e $$forall Ointau_pi,O=bigcup_{iin I}prod_{k=1}^n O_{k,i}$$ (where $O_{k,i}$ is an open set in $tau_k$)$$Rightarrow f^{-1}(O)intau_Y$$ but that's seems fishy to me. Am I (even not completely) right?
general-topology continuity
general-topology continuity
edited Sep 7 '14 at 12:13
Gerry Myerson
146k8147298
146k8147298
asked Sep 7 '14 at 10:29
user65985
$begingroup$
Each open set in the product topology is the union of products of open sets. So you're close to the target.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 10:33
$begingroup$
I added union before the product. Now that's correct?
$endgroup$
– user65985
Sep 7 '14 at 10:36
$begingroup$
You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 11:48
add a comment |
$begingroup$
Each open set in the product topology is the union of products of open sets. So you're close to the target.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 10:33
$begingroup$
I added union before the product. Now that's correct?
$endgroup$
– user65985
Sep 7 '14 at 10:36
$begingroup$
You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 11:48
$begingroup$
Each open set in the product topology is the union of products of open sets. So you're close to the target.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 10:33
$begingroup$
Each open set in the product topology is the union of products of open sets. So you're close to the target.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 10:33
$begingroup$
I added union before the product. Now that's correct?
$endgroup$
– user65985
Sep 7 '14 at 10:36
$begingroup$
I added union before the product. Now that's correct?
$endgroup$
– user65985
Sep 7 '14 at 10:36
$begingroup$
You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 11:48
$begingroup$
You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 11:48
add a comment |
1 Answer
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$begingroup$
Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.
Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$
Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.
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add a comment |
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$begingroup$
Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.
Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$
Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.
$endgroup$
add a comment |
$begingroup$
Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.
Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$
Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.
$endgroup$
add a comment |
$begingroup$
Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.
Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$
Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.
$endgroup$
Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.
Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$
Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.
edited Dec 4 '18 at 9:51
answered Sep 7 '14 at 11:49
Henno BrandsmaHenno Brandsma
106k347114
106k347114
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$begingroup$
Each open set in the product topology is the union of products of open sets. So you're close to the target.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 10:33
$begingroup$
I added union before the product. Now that's correct?
$endgroup$
– user65985
Sep 7 '14 at 10:36
$begingroup$
You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
$endgroup$
– Daniel Fischer♦
Sep 7 '14 at 11:48