Let $X, Y, Z$ be three lctvs. Then https://ncatlab.org/nlab/show/inductive+tensor+product , the first theorem in section 3, tells us that $$\operatorname{Hom}(X\otimes_{\iota} Y, Z) = \operatorname{Hom}(X, L(Y, Z)_{\sigma}),$$
where $L(Y, Z)_{\sigma}$ is the space of continuous linear maps $Y\to Z$ given the topology of pointwise convergence.
Is there some topology on $L(Y,Z)$ I can use so that this holds with the inductive tensor product replaced by the projective tensor product?
Assuming the homomorphisms of locally convex topological vector spaces are taken to be the continuous linear mappings, there are a few situations in which there is no locally convex topology on $L(Y,Z)$ compatible with the vector space structure for which $$\underset{\mathbf{lctvs}}{\operatorname{Hom}}(X\otimes_\pi Y,Z)=\underset{\mathbf{lctvs}}{\operatorname{Hom}}(X,L(Y,Z))$$ holds via the usual linear mapping $$\begin{array}{ccc}\underset{\mathbf{lctvs}}{\operatorname{Hom}}(X\otimes_\pi Y,Z)&\longrightarrow&\underset{\mathbf{Vect}}{\operatorname{Hom}}(X,L(Y,Z))\newline u&\longmapsto&x\mapsto u(x\otimes\cdot~)\end{array}.$$
For example, when $Y$ isn't seminormable and $Z$ is the scalar field, there is always a $X$ for which no good topology exists. More precisely, I show below the following result :
Let $Y$ be a locally convex space. Suppose that there exists a locally convex topology $\mathscr T$ on $Y'$ compatible with the vector space structure such that for every locally convex space $X$ and when $Y'$ is endowed with $\mathscr T$, the linear mapping $$\begin{array}{ccc}(X\otimes_\pi Y)'&\longrightarrow&\underset{\mathbf{Vect}}{\operatorname{Hom}}(X,Y')\newline u&\longmapsto&x\mapsto u(x\otimes\cdot~)\end{array}$$ realises a bijection between $(X\otimes_\pi Y)'$ and $L(X,Y')$. Then $Y$ is seminormable.
The plan is as follows : since the topology of $Y$ is the topology of uniform convergence on the convex balanced equicontinuous subsets of $Y'$, it suffices to show the existence of a convex balanced equicontinuous subsets of $Y'$ which absorbs every other convex balanced equicontinuous subset of $Y'$, as the gauge of the polar of such a set is a seminorm which defines the topology of $Y$.
For convenience, $Y'$ will from now on be endowed with $\mathscr T$ and for every convex balanced equicontinuous subset of $Y'$, I'll denote by $Y'_M$ the seminormable space obtained by endowing $\operatorname{span}(M)$ with the topology defined by the gauge of $M$. I'll denote by $X$ the locally convex space obtained by endowing $Y'$ with the finest locally convex topology such that for every convex balanced equicontinuous subset $M$ of $Y'$, the injection from $Y'_M$ to $X$ is continuous.
Notice that given a locally convex space $W$, a linear form $u$ on $W\otimes_\pi Y$ is continuous if and only if the bilinear form $(x,y)\mapsto u(x\otimes y)$ on $W\times Y$ is continuous, i.e. if and only if there is a neighbourhood $U$ of $0$ in $W$ and a neighbourhood $V$ of $0$ in $Y$ such that $|u(x\otimes y)|\leq 1$ for every $(x,y)\in U\times V$ or equivalently, if and only if there is a neighbourhood $U$ of $0$ in $W$ such that $\lbrace u(x\otimes\cdot~)~|~x\in U\rbrace$ is an equicontinuous subset of $Y'$. So by the assumptions on $\mathscr T$, a linear mapping $v$ from $W$ to $Y'$ is continuous if and only if there is a neighbourhood $U$ of $0$ in $W$ such that $v(U)$ is equicontinuous.
Then, for every convex balanced equicontinuous subset $M$ of $Y'$, the injection from $Y'_M$ to $Y'$ is continuous because $M$ is a neighbourhood of $0$ in $Y'_M$. It follows that the topology of $X$ is finer than that of $Y'$ and that the identity mapping from $X$ to $Y'$ is continuous. So there is a neighbourhood $U$ of $0$ in $X$ which is an equicontinuous subset of $Y'$.
Now, any convex balanced equicontinuous subset $M$ of $Y'$ is bounded in $Y'_M$ and so is bounded in $X$ because the injection from $Y'_M$ to $X$ is continuous. So $U$ absorbs every convex balanced equicontinuous subset of $Y'$.
