Let $\left(\{A_{i}\}_{i\in I},\{\mu_{ij}:A_{i}\to A_{j}|i\leq j\} \right)$ be a direct system of topological spaces. Then for any topological space $B$, $\left(\{A_{i}\times B\}_{i\in I},\{\mu_{ij}\times 1_{B}:A_{i}\times B\to A_{j}\times B|i\leq j\} \right)$ is also a direct system. Then, I wanted to know whether $\varinjlim\left( A_{i}\times B\right)\cong \left(\varinjlim A_{i}\right)\times B$ (homeomorphic) in this case.
Basically, what I tried was to show that $\left(\varinjlim A_{i}\right)\times B$ satisfies the characteristic property:
Whenever $X$ is a topological space and $\forall i\in I,\alpha_{i}:A_{i}\times B\to X$ is a continuous map satisfying $\forall i\leq j, \alpha_{i}=\alpha_{j}\circ \left(\mu_{ij}\times 1_{B}\right)$, then there exists a unique continuous map $\alpha:\left(\varinjlim A_{i}\right)\times B\to X$ such that $\forall i,\alpha_{i}=\alpha\circ\left(\mu_{i}\times 1_{B}\right)$
(Here, $\mu_{i}:A_{i}\to \varinjlim A_{i}$ is the natural map.)
I proved the existence of the set-theoretic map $\alpha$ and so I would be done if I can show that it is continuous. So let $U$ be open in $X$. From $\forall i,\alpha_{i}=\alpha\circ\left(\mu_{i}\times 1_{B}\right)$, we get that $$\alpha^{-1}(U)=\bigcup_{i\in I}\left(\mu_{i}\times 1_{B}\right)\left(\alpha_{i}^{-1}(U)\right).$$
As, $\alpha_{i}$ is continuous, $\alpha_{i}^{-1}(U)$ is open in $A_{i}\times B$ and so if I can show that $\mu_{i}\times 1_{B}$ is an open map, we would be done. Now $1_{B}$ is an open map and so I would be done if I can show that $\mu_{i}$ is an open map.
Now let $V$ be open in $A_{i}$. As we have $$\mu_{i}=A_{i}\hookrightarrow \coprod_{j\in I}A_{j} \twoheadrightarrow \left(\coprod_{j\in I}A_{j}\right)/\sim,$$ we get that $\mu_{i}(V)$ is open if and only if for every $j\in I$, $$\bigcup_{k\geq j}\mu_{jk}^{-1}\left(\mu_{ik}(V)\right)$$ is open in $A_{j}$.
Thus we are done if all the transition maps $\mu_{ij}:A_{i}\to A_{j}$ are open maps. This is where I am stuck.
How do I salvage the proof for the general case? Or is the result not true in general?
It is at least true for locally compact $B$. You certainly know the exponential law which states that there is bijection between continuous maps $f : X \times Y \to Z$ and continuous maps $f' : X \to Z^Y$ (where $Z^Y$ is the space of continuous functions $Y \to Z$ with the compact-open topology) provided $Y$ is locally compact. This bijection is given by $e(f)(x) = f(x,-) : Y \to Z$.
Now you see that the $\alpha_i$ correspond to $\alpha_i' : A_i \to X^B$. We have $\alpha_i' = \alpha_j' \circ \mu_{ij}$, hence there exists a unique continuous $\alpha' : \varinjlim A_{i} \to X^B$ such that $\alpha_i' = \alpha' \circ \mu_i$. Then the adjoint $\alpha : \varinjlim A_{i} \times B \to X$ will do.