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From Intro to Topological Manifolds:

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I want to show that the subspace topology and the product topology on $S_1 \times \cdots \times S_n \subset X_1 \times \cdots \times X_n$ are equal. One way I can think of doing this is to show that the subspace topology satisfies the characteristic property of the product topology. Since the product topology is the unique topology satisfying the characteristic property, then the subspace topology and the product topology will be equal.

(I could also show the product topology satisfies the characteristic property of the subspace topology as well.)

However, what exactly does it mean for a topology to satisfy the characteristic property and how would I begin to show this?

Martin Sleziak
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user5826
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  • your $f$ is the inclusion map. Moreover consider the maps $p_I:\prod_i S_i \rightarrow S_i$ which is restriction of $\pi_i$ and the inclusions $\iota_i:S_i \rightarrow X_i $. Clearly $f\circ\pi_I=p_i\circ \iota_i$. If $\prod_i S_i$ is endowed with the product topology then each $pi_i$ is continuous and also $\iota_i$. Hence the composition are all continuous. From this you can prove easily the theorem. – Dog_69 May 23 '18 at 21:37
  • It's all in great detail and abstraction in my answer here; it also does the subspace of products via universal properties, for products of any size. – Henno Brandsma May 23 '18 at 21:52

1 Answers1

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Let $\iota:S_1\times\cdots\times S_n\to X_1\times\cdots\times X_n$ denote the inclusion function.

If $S_1\times\cdots\times S_n$ is equipped with product topology then its projections are the functions $\pi_i\circ\iota$.

$S_1\times\cdots\times S_n$ is equipped with the subspace topology if and only if for every function $f:Y\to S_1\times\cdots\times S_n$ we have:$$f\text{ is continuous }\iff\iota\circ f\text{ is continuous }\tag1$$

$S_1\times\cdots\times S_n$ is equipped with the product topology if and only if for every function $f:Y\to S_1\times\cdots\times S_n$ we have:$$f\text{ is continuous }\iff\pi_i\circ\iota\circ f\text{ is continuous for every }i\tag2$$

Now observe that the RHS in $(1)$ and $(2)$ is exactly the same condition because a function $g:Y\to X_1\times\cdots\times X_n$ is continuous if and only if $\pi_i\circ g$ is continuous for every $i$. This can be applied on $g=\iota\circ f$.

We conclude that product topology and subspace topology on $S_1\times\cdots\times S_n$ coincide.

drhab
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