Weak topology and subspaces Let $(X_i)_{i \in \mathbb{N}}$ be a family of topological spaces with inclusion maps $j_i: X_i \hookrightarrow X_{i+1}$ (i.e. $X_i$ has subspace topology with respect to this maps). We can take their direct limit (in the category of topological spaces) 
$$
X = \varinjlim_{i \in \mathbb{N}} X_i
$$
which is the "union" of all $X_i$ equipped with the "weak topology", i.e. a set $A \subset X$ is open iff its "intersection" with each $X_i$ is open in $X_i$.
Now we can choose a sequence of subspaces $Y_i \subset X_i$ such that $Y_i \subset Y_{i+1}$ (or more precisely, $j_i(Y_i) \subset Y_{i+1}$). These inclusions give us a morphism between the directed systems $(X_i)$ and $(Y_i)$, so we get a continuous map between their direct limits
$$
Y = \varinjlim_{i \in \mathbb{N}} Y_i \to \varinjlim_{i \in \mathbb{N}} X_i  = X
$$
Clearly this map is injective.
Question: Does $Y$ carry the subspace topology with respect to this map?
 A: I don’t have at hand the definition of the limit map between direct the limits, but I guess the question can have a negative answer. Let $X=\Bbb R^\omega$ be a subspace of a Tychonoff product $\Bbb R^\omega$ consisting of all sequences $x=(x_i)$ such that all but finitely many $x_n$ are zeroes. Then the space $X$ is a direct limit of a sequence of spaces $X_i$, where each $X_i=X$ and the inclusion maps are the identity maps. For each natural $i$ let $Y_i=\{(x_n)\in\Bbb R^\omega: x_n=0 \mbox{ for all }n>i  \}$ and the inclusion maps are the embeddings. Endow the set $Y=\bigcup Y_i=\Bbb R^\omega$ with the topology consisting of all subsets $U$ of $\Bbb R^\omega$ such that $U\cap Y_i$ is open in $Y_i$ for each $i$. That is, $Y$ is a direct limit of the sequence $\{Y_i\}$. Now let $a^n=(a^n_i)$ be a sequence of elements of $\Bbb R^\omega$ such that for each $n$ and $i$ we have $a^n_i$ equals $1$, if $n=i$, and equals $0$, otherwise. Then the sequence $(a^n)$ converges to $0$ in $X$, but not in $Y$.
