Reflexive closure of Banach space Given a Banach space $E$, need there exist a reflexive Banach space $\overline{E}$ and a map $T: E \to \overline{E}$ such that any map $S: E \to X$, where $X$ is a reflexive Banach space, factors through $\overline{E}$ via $T$? This $\overline{E}$ would then be a sort of reflexive "closure" or "envelope" of $E$. 
My initial thought was to look at the colimit of 
$$E \hookrightarrow E^{**} \hookrightarrow E^{****} \hookrightarrow \dots$$
but this is just a wild guess. For it to work, I'd need to know that the double dual commutes with colimits in the category of Banach spaces, and that seems doubtful to me.
 A: The double dual does not commute with colimits as colimits of $C(K)$-spaces are also $C(K)$ but they are never reflexive unless finite-dimensional.

In general, $\overline{E}$ and $T$ with the properties you want need not exist at least if $E$ and $\overline{E}$ are to be separable. 

Proof. Let $E$ be Pełczyński's universal space; it is separable and has the property that every separable Banach space with the bounded approximation property embeds as a complemented subspace thereof. 
However, there exist reflexive spaces $(E_\alpha)_{\alpha<\omega_1}$ with BAP that have arbitrarily large countable Szlenk index, i.e., ${\rm Sz}\, E_\alpha > \omega^\alpha$ (already the spaces constructed by Szlenk have BAP). Let $S_\alpha$ be a projection from $E$ onto a copy of $E_\alpha$ in $E$. If there were $\overline{E}$ and $T$ with said properties, $T$ would factor all $S_\alpha$ so $\overline{E}$ would have to contain complemented subspaces isomorphic to $E_\alpha$ for all $\alpha<\omega_1$. This is impossible as separable reflexive spaces have countable Szlenk index (that is they cannot contain subspaces with arbitrarily large Szlenk index). $\square$

There is however a silly way of producing the pair you want but it can hardly be called a closure. 

Let $E$ be a Banach space and denote by $\lambda$ the density of $E$. Every Banach space of density at most $\lambda$ embeds into $\ell_\infty(\lambda)$, so every operator $S\colon E\to X$ may be regarded as an operator into $\ell_\infty(\lambda)$. Let $(E_\gamma)_{\gamma\in \Gamma}$ be the family of all reflexive subspaces of $\ell_\infty(\lambda)$. Set $$\overline{E} = (\bigoplus_{\gamma\in \Gamma} E_\gamma)_{\ell_2(\Gamma)},$$ which is reflexive. Suppose that $X$ is reflexive and $S\colon E\to X$. Then $X_0 = \overline{S[E]}$ has density at most $\lambda$, so there exists an isomorphim $U\colon X_0\to E_\gamma$ for some $\gamma$. Then $$S = U^{-1}P_\gamma \iota_\gamma US,$$ where $\iota_\gamma\colon E_\gamma\to \overline{E}$ is the standard embedding and $P_\gamma$ is the projection onto $\gamma^{\rm th}$ coordinate.
