Is the tensor product of two representations a representation?

I am a little bit uncertain about an argumentation showing that a given map of a topological group is somehow obviously continuous. In the following I will rely on the book of Anthony W. Knapp „Lie Groups, Lie Algebras and Cohomology“.

Let us fix some notation and definitions;

Let $G$ be a topological group. A finite-dimensional representation of $G$ is a homomorphism $\pi$ of $G$ into the group of invertible linear maps of a finite-dimensional complex vector space $V$ into itself such that the resulting map of $G \times V$ into $V$ is continuous.

Now I would like to understand the following assertion:

Let $G$ be topological group, and let $\sigma$ and $\rho$ be representations of $G$ on finite-dimensional complex vector spaces $V$ and $W$, respectively. Define a representation $\pi$ of $G$ on $V \otimes_{\mathbb{C}} W$ by
$$\pi(g) = \sigma(g) \otimes \rho(g).$$
Then one can use bases to see that $\pi$ is continuous.

I don’t understand the last sentence; how it is supposed to help me to show continuity of the resulting map $G \times (V \otimes_{\mathbb{C}} W) \to V \otimes_{\mathbb{C}} W$. For fixed $g \in G$ one can argue that $\pi(g)$ is a linear map between finite-dimensional vector spaces and therefore continuous. But what about the parameter $g$?

I have to show that the map $(g,m \otimes n) \mapsto \sigma(g) \otimes \rho(g)(m \otimes n) = \sigma(g)(m) \otimes \rho(g)(n)$ is continuous in $(g,m \otimes n) \in G \times (V \otimes_{\mathbb{C}} W)$ and I am therefore obliged to write it as a composition of continuous maps. I know that $\sigma$ and $\rho$ are continuous but I don’t know nothing about the continuity of $\otimes$ and that is where I fail.

I appreciate any sort of suggestion which helps me to understand this somewhat trivial argumentation.

Solutions Collecting From Web of "Is the tensor product of two representations a representation?"

Lemma. If $V$ is a complex representation of the underlying group of $G$, then $G \times V \to V$, $(g,v) \mapsto gv$ is continuous if and only if the homomorphism $G \to \mathrm{GL}(V)$, $g \mapsto (v \mapsto gv)$ is continuous.

Once you have proven this general and useful Lemma, it is easy to construct tensor products. Namely, if $G$ acts on $V$ and on $W$, then we obtain an action of $G$ on $V \otimes W$ via
$$G \to G \times G \to \mathrm{GL}(V) \times \mathrm{GL}(W) \to \mathrm{GL}(V \otimes W).$$
Thus, we only have to prove that $\mathrm{GL}(V) \times \mathrm{GL}(W) \to \mathrm{GL}(V \otimes W)$ is continuous. But this is even a polynomial map, when you write it down in bases.

I will try to give a partial answer to my own question:

We have to show that a map $$f \colon G \times (V \otimes_{\mathbb{C}} W) \to G, \\ (g,u) \mapsto (\sigma(g) \otimes \rho(g))(u)$$
is continuous, where the domain of $f$ is given the product topology.

Let $\{e_i\}_{i \in \{1,\dotsc,n\}}$ be a basis of $V$ and $\{\tilde{e}_j\}_{j\in \{1,\dotsc,m\}}$ be a basis of $W$, then $\{e_i \otimes \tilde{e}_j\}_{i \in \{1,\dotsc,n\}, \,j \in \{1,\dotsc,m\}}$ is a basis for $V \otimes_{\mathbb{C}} W$. Moreover define the following
\begin{align}
\sigma(g)(e_i) = s^k_i(g) e_k, \quad \rho(g)(\tilde{e}_j) = r^l_j(g) \tilde{e}_l,
\end{align}
where by assumption $s^k_i$ and $r^l_j$ are continuous functions from $G$ to $\mathbb{C}$ for all $i,k \in \{1, \dotsc,n\}$ and $l,j \in \{1, \dotsc,m\}$.
Define projections $$p_g \colon G \times (V \otimes_{\mathbb{C}} W)\to G, \,(g,u) \mapsto g \\ p_u \colon G \times (V \otimes_{\mathbb{C}} W)\to G, \,(g,u) \mapsto u.$$ Then we have for $g \in G$ and $u \in V \otimes_{\mathbb{C}} W$
\begin{align}
f(g,u) &= (\sigma(g) \otimes \rho(g))(a^{ij}e_i \otimes \tilde{e}_j) \\
&= a^{ij}s^k_i(g)r^l_j(g)(e_k \otimes \tilde{e}_l) \\
&= \langle p_u(g,u), e_i \otimes \tilde{e}_j \rangle s^k_i(p_g(g,u))r^l_j(p_g(g,u))(e_k \otimes \tilde{e}_l)
\end{align}

This is a composition of continuous functions. Notice that the projections are defined to be continuous in the given product topology. There is one drawback; I have assumed somehow that we have a inner product space. This means that I am not quite right in my argumentation. I project on the $u$-component of $(g,u)$ and then use the inner product $\langle \cdot , \cdot \rangle$ to obtain the coefficient $a^{ij}$. How to avoid this problem about the existence of an inner product?