I'm self studying Lee's Introduction to Topological Manifolds, and I'm familiarising myself to the characteristic/universal property view of topology. One of the exercises in chapter 3 goes as follows:

Suppose $\,f_1,\,f_2:X \rightarrow \mathbb{R}$ are continuous functions. Let $$(f_1+f_2)(x) = f_1(x) + f_2(x),$$ $$(f_1\cdot f_2)(x) = f_1(x)\cdot f_2(x)$$

Use the characteristic property of the product topology to show that the pontwise sums and products of continuous functions are continuous.

Now, I'm still new to this way of dealing with things, so my attempt, which I think is a bit convoluted, was as follows:

The characteristic property states that if $\prod_{\alpha \in A}X_\alpha$ has the product topology and Y is a topological space, a map $\,f:Y\rightarrow\prod_{\alpha \in A}X_\alpha$ is continuous if and only if each of its component functions $f_i = \pi_\alpha \circ f$ are continuous, where $\pi_\alpha$ is the canonical projection.

The book just showed that if $\,f_1$ and $\,f_2$ are continuous, the product map $\,f_1\times f_2$ is continuous.

We build a commutative diagram as follows:

$$X \overset{(Id)}\rightarrow \Delta \overset{(f_1\times f_2)}\rightarrow \mathbb{R}^2 \overset{(+\, \text{or} \,\cdot)}\rightarrow \mathbb{R} \overset{(\pi = \text{Id})}\rightarrow \mathbb{R}$$ and $$X\overset{(f_1+f_2 \,\text{or}\, f_1 \cdot f_2)} \rightarrow \mathbb{R}$$ where $\Delta$ is the diagonal of $X^2$.

The first function is a composition of continuous functions (either inclusion or sum/multiplication within the reals). By the characteristic property, the component function below must also be continuous.

Is this proof correct and is there a better one?

  • 3
    $\begingroup$ You don't need the last $\pi = \operatorname{Id}$, and instead of mapping to the diagonal $\Delta$ to use $f_1\times f_2$, you could directly use $(f_1,f_2)\colon X \to \mathbb{R}^2$. And of course you must show or cite a proposition that addition and multiplication are continuous $\mathbb{R}^2\to\mathbb{R}$. $\endgroup$ – Daniel Fischer May 23 '14 at 22:43

The space of functions is defined as: $$\mathcal{F}(X,\mathbb{R})=\prod_{x\in X}\mathbb{R}$$ and its cartesian product is identifyable by*: $$\mathcal{F}(X,\mathbb{R})\times\mathcal{F}(X,\mathbb{R})=\prod_{x\in X}\mathbb{R}^2$$ So pointwise addition is continuous: $$\oplus:\mathcal{F}(X,\mathbb{R})\times\mathcal{F}(X,\mathbb{R})\to\mathcal{F}(X,\mathbb{R}):(f\oplus g)(x):=f(x)+g(x)$$ The proof exploits the diagram:
Pointwise Operations

*This is actually the only point one needs to prove explicitely: $$\mathcal{F}(X,\mathbb{R})\times\mathcal{F}(X,\mathbb{R})=\prod_{1,2}(\prod_{x\in X}\mathbb{R})\cong\prod_{x\in X}(\prod_{1,2}\mathbb{R})=\mathcal{F}(X,\mathbb{R}^2)$$ Note that these spaces are really different - even from the set-theoretic point of view: $$(i,(x,r))\in\prod_{1,2}(\prod_{x\in X}\mathbb{R}) \qquad (x,(i,r))\in\prod_{x\in X}(\prod_{1,2}\mathbb{R})$$

  • $\begingroup$ Could you clarify how you iterated over $X$ to make the space of functions? What I understand from the notation is that an arbitrary element of $\mathcal{F}(X,\mathbb{R})$ is a pair $( \mathbb{R}, x)$ for $x \in X$, which doesn't define a function from $\,X$ to $\mathbb{R}$. $\endgroup$ – Felipe Jacob May 24 '14 at 10:46
  • $\begingroup$ An arbitrary element of the product $\prod_{i\in I}X_i$ is basically a relation $R\subseteq I\times\cup_iX_i$ with its pairs $(i,x)$ s.t. $x\in X_i$. For the special case $X_i\equiv X$ this just reduces to functions $f:I\to X$ - remember functions are nothing but special relations $(i,f(i))\in I\times X$. $\endgroup$ – C-Star-W-Star May 24 '14 at 17:08

I think the key here is showing $f(x) = (f_1(x),f_2(x))$ is continuous if $f_1$ and $f_2$ are. You'll want to invoke the characteristic property for both canonical projections:

enter image description here

(Note that this is analogous to Lee's proof for the product map $f_1 \times f_2$, but our $f$ is different from this product map.) Combine this with the continuity of $x+y$ and $x\cdot y$ as maps $\mathbb{R}^2 \rightarrow \mathbb{R}$ and $\mathbb{C}^2 \rightarrow \mathbb{C}$ and you'll get to your result.

  • $\begingroup$ Can you elaborate on how this shows the pointwise sum is continuous? How does $f(x) = (f_1(x),f_2(x))$ imply the sum is continuous? $\endgroup$ – Al Jebr May 24 '18 at 16:37
  • $\begingroup$ Continuity of $f_1$ and $f_2$ implies continuity of $f(x) = (f_1(x),f_2(x))$ by the characteristic property. The sum function $s(x,y) = x + y$ is continuous because the pre-image of an open set is open (around any point $(x_0,y_0)$ there is an open ball where $|s(x,y) - s(x_0,y_0)| < \epsilon$). Since $f$ and $s$ are continuous, so is $s \circ f$ which is the result. $\endgroup$ – orlandpm May 24 '18 at 20:21

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