Let $X$ be the union of the coordinate axes in the product set $\Bbb R ^{\Bbb N}$, that is, the set of those points $x$, for which $x_n \ne 0$ for at most one $n \in \Bbb N$. For each $i \in \Bbb N$ we define a map $f_i: \Bbb R \to X$ as follows: $$ f_i(x)_j=\begin{cases} x, \ i=j \\ 0, \ i \ne j \end{cases}$$ Prove, that the topology of $X$, which is coinduced by the maps $f_i$ from the usual topology of $\Bbb R$ is strictly finer than the subspace topology of $X$, which is inherited from the product space $\Bbb R^{\Bbb N}$.
Denote the coinduced/final topology as $\tau_i$. We are to show that $\tau_X \subset \tau_i$ where $\tau_X$ is the subspace topology of $X \subset \Bbb R^{\Bbb N}$.
Consider a basic open set $O \in \tau_X$. The set $O$ is of form $O= \left(\bigcap_{j \in K} \pi_j^{-1}(V_j) \right) \cap X$ for $K \subset J$ finite and $V_j$ open in $\Bbb R$.
Now if we can show that $$O= \left(\bigcap_{j \in K} \pi_j^{-1}(V_j) \right) \cap X \in \tau_i$$ we are done. We have that $O= \left(\bigcap_{j \in K} \pi_j^{-1}(V_j) \right) \cap X \in \tau_i$ if $$f_i^{-1}(O)_j=f_i^{-1} \left( \left(\bigcap_{j \in K} \pi_j^{-1}(V_j) \right) \cap X \right)_j = \bigcap_{j \in K} f_i( \pi_j^{-1}(V_j))_j \cap f_i^{-1}(X)_j$$ is open in $\Bbb R$.
Here I'm stuck I don't know how to show that $$f_i( \pi_j^{-1}(V_j))_j \cap f_i^{-1}(X)_j$$ is open in $\Bbb R$? The result would follow since I'm taking a finite intersection of open sets.
I think it may help to think at a slightly more abstract level. $\tau_i$ is the finest topology on $X$ that makes every $f_i : \Bbb{R} \to X$ continuous (under the usual topology on $\Bbb{R}$), i.e., it comprises all sets $A$ such that for every $i$, $f_i^{-1}[A]$ is open in $\Bbb{R}$. Since every $f_i$ is continuous w.r.t. $\tau_X$, then we must have $\tau_X \subseteq \tau_i$. Do you see why?. So $\tau_i$ is finer than $\tau_X$. To show that $\tau_i$ is strictly finer than $\tau_X$, we need to find a $\tau_i$-open set that is not $\tau_X$-open: to get such a set, pick $\epsilon > 0$ and consider $A = (-\epsilon, \epsilon)^{\Bbb{N}} \cap X$. $A$ is certainly in $\tau_i$. Is it in $\tau_X$?.