Continuity of sum/product using characteristic property of product topology 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?
 A: 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:

(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.
A: 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:

*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:
$$(i,(x,r))\in\prod_{1,2}(\prod_{x\in X}\mathbb{R}) \qquad (x,(i,r))\in\prod_{x\in X}(\prod_{1,2}\mathbb{R})$$
