Injective Connecting Maps in Direct Limits of TVS I am reading Jarchow's Locally Convex Spaces, and I've come across a few questions about direct limits of TVS I have not been able to answer (these questions are based on section 4.5 of Jarchow's book). I start by setting up some notation: let $J$ be a directed set, and for every $j \in J$, let $E_j$ be a TVS. Whenever $j \leq k$, let $S_{kj}:E_j \to E_k$ be a continuous linear map, and suppose that $S_{ki} = S_{kj} \circ S_{ji}$ whenever $i \leq j \leq k$, and suppose $S_{jj}$ is the identity map $E_j \to E_j$. For $i \in J$, let $I_i: E_i \to \bigoplus_{j \in J} E_j$ denote the canonical injection. Let $L: = \text{span} \bigcup_{j \leq k} (I_j - I_k \circ S_{kj})(E_j) $. Let then $\varinjlim E_j := \bigoplus_{j \in J} E_j / L$, and equip $\varinjlim E_j$ with the finest linear topology making the linear maps $S_i:E_i \to \varinjlim E$ continuous, where we define $S_i := Q \circ I_i$, where $Q: \bigoplus E_j \to \varinjlim E_j$ is the quotient map. Say that the direct system $(E_j, S_{kj})$ is reduced if the maps $S_j$ are injective. 
Question 1 
Jarchow states and sketches a proof that any direct limit of TVS is linearly homeomorphic to the direct limit of a reduced system. To do this, define $F_j := E_j /{\text{ker}~ S_j}$, equipped with the quotient topology (which is a linear topology). Then define, for $j \leq k$, $T_{kj}: F_j \to F_k: x+ \text{ker}~ S_j \to S_{kj}(x)+ \text{ker}~ S_k$, which is well-defined since $S_j = S_k \circ S_{kj}$. The continuity of $T_{kj}$ follows from the continuity of $S_{kj}$ and the universal property for maps out of a quotient space. Also define $L':= \text{span} \bigcup_{j \leq k} (I'_j - I'_k \circ T_{kj})(F_j)$, where $I'_i: F_i \to \bigoplus_j F_j$ is the canonical injection. Let $Q': \bigoplus_j F_j \to \varinjlim F_j$ be the associated quotient map, and let $T_j:= Q' \circ I'_j$ for $j \in J$. 
I am able to show $\varinjlim F_j$ (equipped with the finest linear topology making the $T_j$ continuous) is linearly homeomorphic to $\varinjlim E_j$, however I do not see why the new direct system $(F_j, T_{kj})$ is reduced. That is, it's not clear to me why the $T_j$ are injective. The difficulty seems occur when we follow our noses: if $T_j(x) = 0$, then $I'_j(x) = \sum_{i=1}^n (I'_{j_i} - I'_{k_i} \circ T_{k_i j_i})(x_i)$ for some $j_i \leq k_i$, and $x_i \in F_{j_i}$. And from this it is not clear that we should have $x=0$. 
Another idea: certainly the $T_{kj}$ are injective. So, perhaps it is even true that whenever the connecting maps $S_{kj}$ in a general direct system of TVS are injective, the direct system will be reduced; I, however, have been unable to show this as well as I run into the same problem described above.
[Question 1 is actually algebraic in nature, so all the topological properties of the $F_j$ are not so important for this question].
Question 2 Now suppose $(E_j, S_{kj})$ is a reduced direct system; we keep the notation in the first paragraph. Jarchow makes some remarks along the lines that in this situation, if we identify the $E_j$ with a linear subspace of the limit, then we may view the connecting maps as the inclusion maps and the direct limit as the union of the subspaces $E_j$. For this identification to make sense, we should expect the map $S_j: E_j \to S_j(E_j) \subseteq \varinjlim E_i$ to be a homeomorphism onto its image (when equipping the image with the subspace topology). However, it is not clear to me why $S_j^{-1}$ (defined on $S_j(E_j)$) is continuous, or, equivalently, that $S_j$ is an open map. The difficulty with this seems to be that initial and final (linear) topologies don't play well together; let me elaborate. 
First, $S_j = Q \circ I_j$, and if $U$ is a 0 neighbourhood in $E_j$, then $I_j(U)$ is a 0 neighbourhood in $I_j(E_j)$ (equipped with subspace topology induced from the direct sum topology on $\bigoplus_j E_j$). Let us write $I_j(U) = V \cap I_j(E_j)$, where $V$ is some 0 neighbourhood in the direct sum topology. Although $Q$ is an open map, $V \cap I_j(E_j)$, of course, will generally not be a neighbourhood of 0 w.r.t. the direct sum topology, so we cannot say that $Q(V \cap I_j(E_j))$ will be a 0 neighbourhood in the direct limit topology. But we can ask if $Q(V \cap I_j(E_j)) $ is a 0 neighbourhood in $S_j(E_j)$ (that is, there is a 0 neighbourhood $W$ in the direct limit topology such that $W \cap S_j(E_j) = Q(V \cap I_j(E_j))$). However, I have had a hard time showing this, so any hints or ways to proceed would be appreciated. 
Question 3 This is a bit of a soft question. In the realm of TVS, how much generality is lost by restricting attention to direct limits described by unions of subspaces of a fixed vector space? To me, it does not seem much generality is lost: suppose $(E_j, S_{kj})$ is a reduced direct system. (There is no loss in generality in considering reduced systems, as mentioned in Question 1). If it is true that the maps $S_j$ are homeomorphisms onto their image (see Question 2), then I am able to show the following:


*

*As sets, $\varinjlim E_j = \bigcup_{j \in J} S_j(E_j)$, and $S_j(E_j) \subseteq S_k(E_k)$ whenever $j \leq k$. 

*Suppose we let $\tilde S_{kj}: S_j(E_j) \to S_k(E_k)$ be the inclusion map whenever $j \leq k$. Here, each $S_j(E_j)$ has the subspace topology $\tau|_{S_j(E_j)}$ w.r.t. $\tau$, where $\tau$ is the direct limit topology on the limit $\varinjlim E_j$ of the direct system $(E_j, S_{kj})$.  Then, the TVS $(\varinjlim E_j, \tau)$ is linearly homeomorphic to the limit of the direct system $(S_j(E_j), \tilde S_{kj})$. Moreover, $\tau$ is the same as the finest linear topology on $\bigcup_{i \in J} S_i(E_i)$ w.r.t. the inclusion maps $S_j(E_j) \to \bigcup_{i \in J} S_i(E_i)$. 


In particular, identifying $E_j$ with its image $S_j(E_j)$, it seems that we may as well think of direct limits as unions of subspaces of a vector space. That is, it seems that we don't lose much if we assume from the outset that the $S_{kj}$ are the inclusion maps, where the $E_j$ are subspaces of some vector space $E$. 
Finally, I'd be happy to post this as three separate questions instead of one since they are each a bit lengthy and I'm not sure on the precise policy regarding posts such as this. (And all three questions are related, especially Questions 2 and 3). 
 A: This is not a full answer, but rather an answer to Question 2 in light of Jochen's comment (I have yet to figure out Question 1). Before presenting a counterexample I found based on Jochen's comment, I begin with some generalities, which also serve the purpose of collecting some basic results with proofs. Numbers (1), (2), and (5) of the following lemma are based on remarks from Bourbaki on locally convex direct limits (though convexity assumptions are unnecessary), and the others are observations. 
Lemma. Suppose $(J, \leq)$ is directed, and suppose to every $j \in J$, $(E_j, \tau_j)$ is a TVS. For $j \leq k$, let $S_{kj}: E_j \to E_k$ be a continuous linear map. Let $S_j: E_j \to \varinjlim E_j$ be the canonical injections. Denote by $\tau_{\ell}$ the direct limit topology on $\varinjlim E_j$. Then, 


*

*$S_j(E_j) \subseteq S_k(E_k)$ whenever $j \leq k$, and $\varinjlim E_j = \bigcup_j S_j(E_j)$ as sets [this is stated in my question under the pretense that the $S_j$ need to be injective, but this is not needed]. 

*On $S_j(E_j)$, let $\tau_j'$ denote the finest linear topology on $S_j(E_j)$ making the map $S_j:E_j \to S_j(E_j)$ continuous. Then, $\tau_{\ell}$ is the same as the finest linear topology $\tau_{\ell}'$ on $\bigcup_{i} S_i(E_i)$ making the linear maps $\iota_j: (S_j(E_j), \tau_j') \to \bigcup_{i} S_i(E_i)$ continuous. 

*In the notation of (2), it is also true that $\tau_{\ell}$ is the same as the the finest linear topology $\tau_{\ell}''$ making the linear maps $\kappa_j: (S_j(E_j), \tau_{\ell} |_{S_j(E_j)} ) \to \bigcup_{i} S_i(E_i)$ continuous, where $\tau_{\ell}|_{S_j(E_j)}$ is the topology induced on $S_j(E_j)$ from $\tau_{\ell}$. 

*If the direct system $(E_j, S_{kj})$ is reduced, the $S_j$ are homeomorphims onto their image as maps $(E_j, \tau_j) \to (S_j(E_j), \tau'_j)$, although as the counterexample below shows, the $S_j$ need not be homeomorphisms $(E_j, \tau_j) \to (S_j(E_j), \tau_{\ell}|_{S_j(E_j)})$.

*Assume the direct system $(E_j, S_{kj})$ is reduced. For $j \leq k$ let $\tilde S_{kj}: (S_j(E_j), \tau_j') \to (S_k(E_k), \tau_k')$ be the inclusion map; this map is continuous. Moreover, if $\varinjlim S_j(E_j)$ denotes the limit of the direct system $(S_j(E_j), \tilde S_{kj})$, then $\varinjlim E_j$ and $\varinjlim S_j(E_j)$ are linearly homeomorphic.  

*$\tau_{\ell}|_{S_j(E_j)} = \tau_j' $ iff $S_j$ is a homeomorphism $(E_j, \tau_j) \to (S_j(E_j), \tau_{\ell}|_{S_j(E_j)})$. 

*Suppose now that all the $E_j$ are subspaces of some fixed vector space and $E_j \subseteq E_k$ whenever $j \leq k$. Let $S_{kj}$ be the inclusion maps. Then, $\varinjlim E_j$ is linearly homeomorphic to $\bigcup_j E_j$, equipped with the finest linear topology with respect to the inclusions $\iota_j: E_j \to \bigcup_i E_i$. Further, the $S_j$ are homeomorphisms $(E_j, \tau_j) \to (S_j(E_j), \tau_{\ell}|_{S_j(E_j)})$ iff the $\iota_j$ are homeomorphisms onto their image. 


Proof. 


*

*Just note $S_j = S_k \circ S_{kj}$ for the first statement, and the second statement follows by writing $(x_j)_j + L = \sum_{i=1}^n (I_{j_i}(x_{j_i})+L) = \sum_{i=1}^n S_{j_i}(x_{j_i}) = \sum_{i=1}^n S_l(y_i) \in S_l(E_l)$, where $L$ is the subspace defined in the introductory paragraph of my question, $\{j_1, \ldots, j_n\} = \{j | x_j \neq 0\}$, $l$ satisfies $j_1, \ldots, j_n \leq l$, and $y_i \in E_l$ are chosen so that $S_l(y_i) = S_{j_i}(x_{j_i})$, in view of the first statement.

*By writing $S_j = \iota_j \circ S_j$, one sees that the $S_j$ are continuous with respect to $\tau_{\ell}'$, so $\tau_{\ell}' \subseteq \tau_{\ell}$. Conversely, by the definition of $\tau_j'$, $\iota_j: (S_j(E_j), \tau_j') \to (\bigcup_i S_i(E_i) , \tau_{\ell} )$ is continuous iff $\iota_j \circ S_j = S_j$ is continuous, and hence the reverse inclusion follows. 

*The definition of subspace topology implies the $\kappa_j$ are continuous with respect to $\tau_{\ell}$, so $\tau_{\ell}'' \supseteq \tau_{\ell}$. To see that the $S_j$ are continuous with respect to $\tau_{\ell}''$, just consider the composition $\kappa_j \circ S_j: (E_j, \tau_j) \to (S_j(E_j), \tau_{\ell}|_{S_j}(E_j) \to (\bigcup_i S_i(E_i), \tau_{\ell}'')$. Whence the other inclusion since $\tau_{\ell}$ is the finest linear topology on $\bigcup_j S_j(E_j)$ making the $S_j$ continuous. 

*The only nontrivial point is continuity of $S_j^{-1}: (S_j(E_j), \tau_j') \to (E_j, \tau_j)$; but by the universal property of final linear topologies, $S_j^{-1}$ is continuous iff $S_j^{-1} \circ S_j$ is continuous, which is evident. 

*Using (4), the map $T: \bigoplus E_j \to \bigoplus S_j(E_j): (x_j)_j \mapsto (S_j(x_j))_j$ is a linear homeomorphism (when equipping each factor in the second direct sum with the topology $\tau_j'$). This map passes to a well-defined continuous map between the quotients, and the inverse map is just as easily constructed. 

*It is always true $\tau_j' \supseteq \tau_{\ell}|_{S_j(E_j)})$, since the map $S_j: (E_j, \tau_j) \to (S_j(E_j), \tau_{\ell}|_{S_j(E_j)}$ is continuous, and $\tau_j'$ is finest linear topology making $S_j$ continuous. The 'only if' direction is trivial, and the 'if' direction follows by (4).

*Let us define the dangerous-looking map $T: \varinjlim E_j \to \bigcup_j S_j(E_j): (x_j)_j + L \mapsto \sum_j x_j $ (only at most finitely many terms of the sum are nonzero). The map is well-defined: if $(x_j-y_j)_j \in L$, then there is $j_1, \ldots j_n, k_1, \ldots, k_n$, with $j_i \leq k_i$, and $z_i \in E_{j_i}$ such that $(x_j-y_j)_j = \sum_{i=1}^n (I_{j_i} - I_{k_i})(z_i)$. Then $\sum_{j \in J} (x_j-y_j) = \sum_{i=1}^n (z_i-z_i) = 0$. And $T$ is continuous because $T \circ S_j = \iota_j$. Now we define the even more dangerous-looking map $S: \bigcup_{j} E_j \to \varinjlim E_j: x \mapsto I_j(x) + L$ if $x \in E_j$. Alas, it is well-defined, for, if $x \in E_j \cap E_k$, choose some $l$ such that $j, k \leq l$. Then $x \in E_l$ also, and we have $I_j(x) - I_l(x), I_k(x) - I_l(x) \in L$ by the definition of $L$, hence $I_j(x)- I_k(x) \in L$. And notice that $S \circ \iota_j = S_j$, so continuity of $S$ follows by the universal property for final linear topologies. That $T \circ S$ is the identity is clear, and to see $S \circ T $ is also the identity, fix a representative $(y_j)_j$ of $(x_j)_j+ L$. Write $\{j_1, \ldots, j_n\} = \{j | y_j \neq 0\}$, and take $l $ such that $j_1, \ldots, j_n \leq l$. Then, $I_l(\sum_j y_j ) - (y_j)_j = \sum_{i=1}^n (I_l - I_{j_i})(y_{j_i}) \in L$; this proves the first claim. The second claim follows by writing $T \circ S_j = \iota_j$. $\square$
Now I present a counterexample. Fix Lebesgue measure on the unit interval $[0,1]$, and consider the Lebesgue-spaces $L^p([0,1])$ with their usual norm $\| \cdot \|_p$. Let $I=(1, \infty)$, directed by $p \preceq q$ iff $q \leq p$. Let $J$ be any countable cofinal subset of $I$. Consider the direct system $(L^p, S_{pq})$, where the $S_{pq}$ are the inclusion maps for $q \preceq p$. By (7) in the Lemma, the direct limit of this system is linearly homeomorphic to $E: = \bigcup_{p \in J} L^p$, equipped with the finest linear topology $\tau_{\ell}$ with respect to the inclusions $\iota_p: L^p \to E$. Suppose to the contrary that all the $\iota_p$ are homeomorphisms onto their image. Fix now $p_0 < q_0,$ where $ p_0, q_0 \in J$. Then we have $\iota_{q_0}(L^{q_0}) = \iota_{p_0}(L^{q_0})$ as sets, and it follows from this trivially that $\tau_{\ell}|_{\iota_{q_0}(L^{q_0})} = \tau_{\ell}|_{\iota_{p_0}(L^{q_0})}$. By the fact that $\iota_{p_0}, \iota_{q_0}$ are homeomorphisms onto their images, it follows that $(L^{q_0}, \| \cdot \|_{q_0}) \cong (\iota_{q_0}(L^{q_0}), \tau_{\ell}|_{\iota_{q_0}(L^{q_0})}) = (\iota_{p_0}(L^{q_0}), \tau_{\ell}|_{\iota_{p_0}(L^{q_0})}) \cong (L^{q_0}, \| \cdot \|_{p_0})$. 
For $0< \varepsilon < 1 - \frac{p_0}{q_0}$, consider the a.e. defined function $f_{\varepsilon}(x) = x^{(\varepsilon-1)/{q_0}}$, $x \in (0,1]$. One then computes $\|f_{\varepsilon}\|_{q_0} = \varepsilon^{-1/{q_0}}$, and $\|f_{\varepsilon}\|_{p_0} \geq (1- p_0/{q_0})^{\frac{1}{p_0}}$. It follows that $\sup_{\|f\|_{p_0} \neq 0} \frac{ \|f \|_{q_0}}{\|f \|_{p_0}}$ is not finite, a contradiction. The result now follows in view of the second claim in (7). 
