For every space $X$, $C_p(X)$ is a topological group. I try to show that for every space $X$, $(C_p(X), +)$ where $$+:C_p(X)\times C_p(X)\to C_p(X):(f,g)\mapsto f+g$$ and for every $x\in X$, $(f+g)(x)=f(x)+g(x)$ is a topological group.
The family $$\{O(f, x_1,\ldots, x_n, \epsilon) : n \in\Bbb N, x_1,\ldots, x_n \in X,\epsilon > 0\}\;,$$ Where $$O(f, x_1,\ldots, x_n,\epsilon) =\{g \in C_p(X) : \vert g(x_i)- f(x_i) \vert<\epsilon \; \text{for all }i\leq n\}$$
is a local base of $C_p(X)$ at $f$. How can we show that the inverse function is continuous?
 A: As Daniel pointed out in his comment, the product space $\Bbb K^X$, which coincides with the space of all functions $X\to \Bbb K$ endowed with the topology of pointwise convergence, forms a topological group via pointwise addition and negation.
$$\require{AMScd}
\begin{CD}
\Bbb K^X×\Bbb K^X @>+>> \Bbb K^X \\
@VVε_x×ε_xV @VVε_xV \\
\Bbb K×\Bbb K @>+>> \Bbb K
\end{CD}$$
To show this, we can use the universal property of the product topology on $\Bbb K^X$ as being the initial topology with respect to all evaluations $\varepsilon_x:f\mapsto f(x)$, where $x$ ranges over the points in $X$. The upper map in the commutative diagram above is thus continuous if $+(ε_x×ε_x)$ is. But this map is continuous since $ε_x$ and $+$ are continuous.
A similar diagram shows that the negation mapping $-:\Bbb K^X \to \Bbb K^X$ is continuous. Note that this method works for any topological group $\Bbb K$, not just for $\Bbb R$.
Now $C(X)$ is a subgroup of $\Bbb K^X$, which endowed with the subspace topology is again a topological group.
