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Let $f_i: \mathbb{R}^n \to \mathbb{R}^{n + 1}$ be defined by $$ f_i(x_1, \ldots, x_n) = (x_1, \ldots, x_{i - 1}, 1,x_i, \dots, x_n). $$ I'm trying to show as explicitly as possible why this is continuous. The image has the subspace topology in $\mathbb{R}^{n + 1}$. Our open sets are intersections of the hyperplane with unions of $n + 1$-dimensional open balls. They sure look like they should be open, but I don't know how to say it other than by saying "the map sends $\mathbb{R}^n$ to an embedding of $\mathbb{R}^n$ in $\mathbb{R}^{n + 1}$ and doesn't 'change' anything, i.e. acts like the identity", which seems hand-wavy.

lanf
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    It should be not too horrible to directly work out the inverse image of an open set. You'll see that they're open. – Randall Feb 24 '21 at 19:32
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    You seem to be checking that open sets in the image subspace have open preimages. That's not the definition of continuity. You should check preimages of open sets in $\mathbb{R}^{n+1}$, not the image. – Randall Feb 24 '21 at 19:33
  • You could say that each component, namely, $x_k$ or $1$ for $k=1,...,n$ is continuous. – Gio Feb 24 '21 at 19:40
  • @Gio I'm not sure that a function which is continuous in each component is always continuous. – lanf Feb 25 '21 at 19:18
  • @Randall I'm a bit stuck here. We start with an open ball in the target space, and we look at it's preimage. If it doesn't intersect with the hyperplane, the preimage is empty, hence open. So assume it does. Then the intersection is an open ball in the subspace/hyperplane. This seems intuitively true from the three-dimensional case, but I don't know how to prove it. – lanf Feb 25 '21 at 19:21
  • An open ball in the hyperplane would have an open preimage, though, since the analogous open ball in $\mathbb{R}^n$ gets sent to it, and the map is injective. – lanf Feb 25 '21 at 23:56

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The universal property of product spaces applied to $\Bbb R^{n+1}$, where $X$ can be any space:

A map $f:X \to \Bbb R^{n+1}$ is continuous iff $\pi_j \circ f: X \to \Bbb R$ is continuous for all $n+1$ projections $\pi_1, \ldots , \pi_{n+1}$ on $\Bbb R^{n+1}$.

Then note that $\pi_j \circ f_i = \pi'_j$ for $j \in \{1, 2, \ldots, i-1\}$ (where $\pi'_j$ are the continuous projections on $X=\Bbb R^n$), and $\pi_i \circ f_i \equiv t$ and a constant map is continuous, and $\pi_j \circ f_i = \pi'_{j-1}$ for $j \in \{i+1, \ldots, n+1\}$, so it immediately follows from the universal property (which holds for all maps into products) that $f_i$ as defined is continuous.

Henno Brandsma
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  • This is certainly clear and explicit, but I was hoping for something which only uses the tools one would have from a first course in point-set topology. – lanf Feb 25 '21 at 23:59
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    @bxw these are exactly the arguments I learnt in my first topology course. – Henno Brandsma Feb 26 '21 at 07:47
  • Okay, fair enough. Could you point me in the direction of a reference for the proof of this universal property? Is it possible to fully understand the definitions of everything here without knowing what a category is? In particular, how can we talk about what a universal property is without category theory? – lanf Feb 27 '21 at 00:20
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    @bxw This post here also gives theory from by first topology course. It's quite elementary (theory of initial and final topologies, (final is another posting of mine)) and just requires one to know about subbases a bit. You can use the names universal theorem/property without doing any category theory proper. – Henno Brandsma Feb 27 '21 at 06:43
  • Thank you, this is very helpful. – lanf Feb 28 '21 at 17:37